Lecture Notes. 15MA301-Probability & Statistics. Prepared by S ATHITHAN

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1 Lecture Notes 5MA3-Probability & Statistics Prepared by S ATHITHAN Assistant Professor Department of of Mathematics Faculty of Engineering and Technology SRM UNIVERSITY Kattankulathur-633, Kancheepuram District. SRM UNIVERSITY Kattankulathur-633, Kancheepuram District.

2 Contents Probability 3. Defintion of Probability and Basic Theorems Random Experiment The Approaches to Define the Probability Conditional Probability and Independents Events Total Probability and Bayes Theorem Discrete and Continuous Distributions 5 3 Mathematical Expectation 5 3. Properties of Expectation Variance and its properties Moments and Moment Generating Function 6 4. Moments Moment Generating Function (MGF) Example Worked out Problems 6 Exercise/Practice/Assignment Problems 33 Page of 36

3 TOPICS: Probability & Statistics Unit- Probability and Random Variables Probability Concepts and Problems Random Variables (Discrete and Continuous) General Probability Distributions and Problems Moments and Moment Generating Functions Page of 36

4 Probability Probability & Statistics. Defintion of Probability and Basic Theorems.. Random Experiment An Experiment whose outcome or result can be predicted with certainty is called a deterministic experiment. An Experiment whose all possible outcomes may be known in advance, the outcome of a particular performance of the experiment cannot be predicted owing to a number of unknown cases is called a random experiment. Definition. (Sample Space). Set of all possible outcomes which are assumed to be equally likely is called the sample space and is usually denoted by S. Any subset A of S containing favorable outcomes is called an event... The Approaches to Define the Probability Definition. (Mathematical or Apriori Definition of Probability). Let S be a sample space and A be an event associated with a random experiment. Let n(s) and n(a) be the number of elements of S and A. Then the probability of event A occurring, denoted by P(A) and defined as P (A) = n(a) n(s) = Number of cases favorable to A Exhaustive number of cases in S(Set of all possible cases) Definition.3 (Statistical or Aposteriori Definition of Probability). Let a random experiment be repeated n times and let an event A occur n times out of n trials. The ratio n is called n the relative frequency of the event A. As n increases, n shows a tendency to stabilize and to n approach a constant value and is denoted by P(A) is called the probability of the event A. i.e., P (A) = lim n n Definition.4 (Axiomatic Approach/Definition of Probability). Let S be a sample space and A be an event associated with a random experiment. Then the probability of the event A is denoted by P(A) is defined as a real number satisfying the following conditions/axioms: (i) P (A) (ii) P(S)= (iii) If A and B are mutually exclusive events, then P (A B) = P (A) + P (B) Axiom (iii) can be extended to arbitrary number of events. i.e., If A, A,..., A n,... be mutually exclusive, then P (A A A 3 A n... ) = P (A ) + P (A ) + P (A ) + P (A 3 ) + + P (A n ) +... n Theorem. The probability of the impossible event is zero. i.e., if φ is a subset of S containing no event, then P (φ) =. Page 3 of 36

5 Theorem. If A be the complimentary event of A, then P (A) = P (A). Theorem 3 (Addition Theorem of Probability). If A and B be any two events, then P (A B) = P (A) + P (B) P (A B) P (A) + P (B). The above Theorem can be extended for 3 events which is given by P (A B C) = P (A) + P (B) + P (C) P (A B) P (B C) P (C A) + P (A B C) Theorem 4. If B A, then P (B) P (A).. Conditional Probability and Independents Events The conditional probability of an event B, assuming that the event A has happened, is denoted by P (B/A) and is defined as P (B/A) = P (A B), provided P (A) P (A) Theorem 5 (Product Theorem of Probability). P (A B) = P (A) P (B/A) The product theorem can be extended to three events such as P (A B C) = P (A) P (B/A) P (C/A and B) There are few properties for the conditional distribution. Please go through it in the textbook. When two events A and B are independent, we have P (B/A) = P (B) and the Product Theorem takes the form P (A B) = P (A) P (B). The converse of this statement is also. i.e., If P (A B) = P (A) P (B), then the events A and B are said to be independent. This result can be extended to any number of events. i.e., If A, A,..., A n be n no. of events, then P (A A A 3 A n ) = P (A ) P (A ) P (A 3 ) P (A n ) Theorem 6. If the events A and B are independent, then the events A and B (and similarly A and B, A and B) are also independent..3 Total Probability and Bayes Theorem Page 4 of 36

6 Theorem 7 (Theorem of Total Probability). If B, B,..., B n be the set of exhaustive and mutually exclusive events and A is another event associated with(or caused by)b i, then P (A) = n P (B i )P (A/B i ) i= Theorem 8 (Bayes Theorem of Probability of Causes). If B, B,..., B n be the set of exhaustive and mutually exclusive events associated with a random experiment and A is another event associated with(or caused by)b i, then P (B i /A) = P (B i)p (A/B i ) n, i =,,..., n P (B i )P (A/B i ) Discrete and Continuous Distributions i= For the problems on Discrete and Continuous Random Variables(RV s) we have to note down the following points from the table for understanding. Operator Discrete RV (Probability Distribution Function) Continuous RV (Probability Density Function) Takes Values of the form (Eg.),, 3,... < x < Standard Notation P (X = x) f(x) CDF F (X = x) = P (X x) F (X = x) = P (X x) Total Probability Σ P (X = x) = f(x)dx = RV-Random Variable, *CDF-Cumulative Distribution Function or Distribution Function 3 Mathematical Expectation Definition 3. (Expectation). An average of a probability distribution of a random variable X is called the expectation or the expected value or mathematical expectation of X and is denoted Page 5 of 36

7 by E(X). Definition 3. (Expectation for Discrete Random Variable). Let X be a discrete random variable taking values x, x, x 3,... with probabilities P (X = x ) = p(x ), P (X = x ) = p(x ), P (X = x 3 ) = p(x 3 ),..., the expected value of X is defined as E(X) = x i p(x i ), i= if the right hand side sum exists. Definition 3.3 (Expectation for Continuous Random Variable). Let X be a continuous random variable with pdf f(x) defined in (, ), then the expected value of X is defined as E(X) = xf(x)dx. Note: E(X) is called the mean of the distribution or mean of X and is denoted by X or µ. 3. Properties of Expectation. If c is any constant, then E(c) = c.. If a, b are constants, then E(aX + b) = ae(x) + b. 3. If (X, Y ) is two dimensional random variable, then E(X + Y ) = E(X) + E(Y ). 4. If (X, Y ) is two dimensional independent random variable, then E(XY ) = E(X)E(Y ). 3. Variance and its properties Definition 3.4. Let X be a random variable with mean E(X), then the variance of X is defined as E(X ) [E(X)] and id denoted by V ar(x) or σ X. Properties of Variance. V ar(ax) = a V ar(x). V ar(ax ± b) = a V ar(x). 4 Moments and Moment Generating Function 4. Moments Definition 4. (Moments about origin). The r th moment of a random variable X about the origin is defined as E(X r ) and is denoted by µ r. Moments about origin are known as raw moments. Page 6 of 36

8 Note:By moments we mean the moments about origin or raw moments. The first four moments about the origin are given by. µ = E(X)=Mean. µ = E(X ) 3. µ 3 = E(X 3 ) 4. µ 4 = E(X 4 ) Note: V ar(x) = E(X ) [E(X)] = µ µ =Second moment - square of the first moment. Definition 4. (Moments about mean or Central moments). The r th moment of a random variable X about the mean µ is defined as E[(X µ) r ] and is denoted by µ r. The first four moments about the mean are given by. µ = E(X µ) = E(X) E(µ) = µ µ =. µ = E[(X µ) ] = V ar(x) 3. µ 3 = E[(X µ) 3 ] 4. µ 4 = E[(X µ) 4 ] Definition 4.3 (Moments about any point a). The r th moment of a random variable X about any point a is defined as E[(X a) r ] and we denote it by m r. The first four moments about a point a are given by. m = E(X a) = E(X) a = µ a. m = E[(X a) ] 3. m 3 = E[(X a) 3 ] 4. m 4 = E[(X a) 4 ] Relation between moments about the mean and moments about any arbitrary point a Let µ r be the r th moment about mean and m r be the r th moment about any point a. Let µ be the mean of X. Page 7 of 36

9 µ r = E[(X µ) r ] = E[(X a) (µ a)] r = E[(X a) m ] r = E [ (X a) r r C (X a) r m + r C (X a) r (m ) + ( ) r (m ) r] = E(X a) r r C E(X a) r m + r C E(X a) r (m ) r C 3 E(X a) r 3 (m ) 3 + r C 4 E(X a) r 4 (m ) 4 + ( ) r (m ) r = m r r C m r m + r C m r (m ) r C 3 m r 3(m ) 3 + r C 4 m r 4(m ) 4 + ( ) r (m ) r We define(fix) m =, then we have µ = m m m = µ = m C m m + (m ) = m (m ) µ 3 = m 3 3 C m m + 3 C m (m ) (m ) 3 = m 3 3m m + (m ) 3 µ 4 = m 4 4 C m 3 m + 4 C m (m ) 4 C 3 m (m ) 3 + (m ) 4 = m 4 4m 3 m + 6m (m ) 3(m ) 4 4. Moment Generating Function (MGF) Definition 4.4 (Moment Generating Function (MGF)). The moment generating function of a random variable X is defined as E(e tx ) for allt (, ). It is denoted by M(t) or M X (t). M X (t) = E(e tx ) = n e tx P (X = x) if X is discrete x= x= e tx f(x)dx if X is continuous If X is a random variable. Its MGF M X (t) is given by ( M X (t) = E(e tx ) = E + tx + (tx) + (tx)3!! 3! + + (tx)r r! ) +... = + t (t) E(X) +!! E(X ) + (t)3 3! E(X3 ) + + (t)r r! E(Xr ) +... = + t! µ + (t)! µ + (t)3 3! µ (t)r r! µ r +... Page 8 of 36

10 i.e. Probability & Statistics µ = Coefft. of t in the expansion of M X (t) µ = Coefft. of t! in the expansion of M X(t). µ r = Coefft. of tr r! in the expansion of M X(t). M X (t) generates all the moments about the origin. (That is why we call it as Moment Generating Function (MGF)). Another phenomenon which we often use to find the moments is given below: We have µ r = M (r) X () M X (t) = E(e tx ) = + t! µ + (t)! µ + (t)3 3! µ (t)r r! µ r +... Differentiate with respect to t, we get M X(t) = µ + tµ + (t)! µ (t)r (r )! µ r +... M M (r) Putting t =, we get X(t) = µ + tµ 3 + (t)! µ (t)r (r )! µ r X (t) = µ r + terms of higher powers of t. M X() = µ M X() = µ. M (r) X () = µ r. The Maclaurin s series expansion given below will give all the moments. M X (t) = M X () + t! M X() + t! M X() +... Page 9 of 36

11 The MGF of X about its mean µ is M X µ (t) = E [ e t(x µ)]. Similarly, the MGF of X about any point a is M X a (t) = E [ e t(x a)]. Properties of MGF. M cx (t) = M X (ct). M X+c (t) = e ct M X (t) 3. M ax+b (t) = e bt M X (at) 4. If X and Y are independent RV s then, M X+Y (t) = M X (t) M Y (t). 5 Example Worked out Problems Example:. From 6 positive and 8 negative numbers, 4 are chosen at random(without replacement) and multiplied. What is the probability that the product is negative? Hints/Solution: If the product is negative(-ve), any one number is -ve (or) any 3 numbers are -ve. No. of ways choosing negative no. = 6 C 3 8 C = 8 = 6. No. of ways choosing 3 negative nos. = 6 C 8 C 3 = 6 56 = 336. Total no. of ways choosing 4 nos. out of 4 nos. = 4 C 4 =. P (the product is -ve) = = 496. Example:. A box contains 6 bad and 4 good apples. Two are drawn out from the box at a time. One of them is tested and found to be good. What is the probability that the other one is also good. Hints/Solution: Let A be an event of first drawn apple is good and let B be an event that the other one is also good. The required probability is P (B/A) = P (A B) P (A) = 4 C / C 4 = 6/45 4/ = 3 Aliter: Probability of the first drawn apple is good= 4. Once the first apple is drawn, the box contains only 9 apples in total and 3 good apples. Probability of the second drawn apple is also good= 3 9 = 3. Page of 36

12 Example: 3. The chances of A, B and C become the general manager of a certain firm is in the ratio 4::3. The probabilities that the bonus scheme will be introduced in the company if A, B and C become general manager is.3,.7 and.8 respectively. Find the probability that the bonus scheme is introduced. If the bonus scheme has been introduced, what is the probability that B has been appointed as general manager?. Hints/Solution: Let B be an event of the bonus scheme is introduced. Let A, A, A 3 be the event of A, B, C becoming the general manager respectively. Given, P (A ) = 4 9, P (A ) = 9, P (A 3) = 3 9 and P (B/A ) =.3 = 3, P (B/A ) =.7 = 7, P (B/A 3) =.8 = 8 By Total Probability Theorem and Bayes Theorem, the required probabilities are n P (B) = P (A i )P (B/A i ) = = 5 9 and i= P (A /B) = P (A )P (B/A ) n = 4/9 5/9 = 4 5 P (A i )P (B/A i ) i= Example: 4. A bag contains 5 balls and if it not known that how many of them are white. Two balls are drawn at random from the bag and they are noted to be white. What is the probability that the bag contains at least white balls and all the balls in the bag are white? Hints/Solution: Let B be an event that two balls are drawn, they turns to be white. i.e., the bag contains at least white balls. Let A, A, A 3, A 4 be the event that the bag contains,3,4,5 white balls respectively. Here we may imagine that there are 4 different bags with w,3w,4w,5w balls. If we choose the bag that may be any one of these bags, so all the events A, A, A 3, A 4 are equally likely. P (A ) = 4, P (A ) = 4, P (A 3) = 4, P (A 4) = 4 and P (B/A ) = C 5 C =, P (B/A ) = 3 C 5 C = 3, P (B/A 3 ) = 4 C 5 C = 6, P (B/A 4) = 5 C 5 C =. By Total Probability Theorem and Bayes Theorem, the required probabilities are n P (B) = P (A i )P (B/A i ) = ( ) = and i= P (A /B) = P (A )P (B/A ) n P (A i )P (B/A i ) i= = /4 / = Page of 36

13 Example: 5. % of women at age forty who participate in routine screening have breast cancer. 8% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer? The same problem can get the form by changing the percentages to real numbers as follows: out of, women at age forty who participate in routine screening have breast cancer. 8 of every women with breast cancer will get a positive mammography. 95 out of 9,9 women without breast cancer will also get a positive mammography. If, women in this age group undergo a routine screening, about what fraction of women with positive mammographies will actually have breast cancer? Hints/Solution: Now we pass both groups through the sieve; note that both sieves are the same; they just behave differently depending on which group is passing through. Let A be an event that a Woman with/having cancer. Let B be an event of showing a positive mammography. Finally, to find the probability that a positive test actually means cancer, we look at those who passed through the sieve with cancer, and divide by all who received a positive test, cancer or not. P(A/B) = P(B/A)P (A) P(B/A)P (A) + P(B/A)P (A) = % 8% (% 8%) + (99% 9.6%) = 7.8% Page of 36

14 All women In measuring, we find: Women with cancer P(A) = % Women without cancer P(A) = 99% P(B/A) =.8 P(B/A) =.96 % 8% 99% 9.6% Example: 6. A random variable X has the following distribution X P[X=x] k k k 3k k k 7k + k Find (i) the value of k,(ii) the Cumulative Distribution Function (CDF) (iii) P (.5 < X < 4.5/X > ) and (iv) the smallest value of α for which P (X α) >. Hints/Solution: (i) We know that Here, P [X = x] =. P [X = ] + P [X = ] + P [X = ] + P [X = 3]+ P [X = 4] + P [X = 5] + P [X = 6] + P [X = 7] = i.e. + k + k + k + 3k + k + k + 7k + k = i.e. k + 9k = i.e. (k )(k + ) = = k = or k = Since the probability is not negative, we take the value k =. Page 3 of 36

15 The probability distribution is given by X P [X = x] k k k 3k k k 7k + k P [X = x] 3 7 (ii) The Cumulative Distribution Function (CDF) F (x) = P (X x) F () = P (X ) = P (X = ) = F () = P (X ) = P (X = ) + P (X = ) = + = F () =. P (X ) = P (X = ) + P (X = ) + P (X = ) = + + = 3 F (7) = P (X 7) = P (X = ) + P (X = ) + P (X = ) + P (X = ) +P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) = = The detailed CDF F (X = x) is given in the table. X P [X = x] k k k 3k k k 7k + k P [X = x] F [X = x] = (iii) P (.5 < X < 4.5/X > ) = P [(.5 < X < 4.5) (X > )] P (X > ) P (.5 < X < 4.5) = P (X = 3) + P (X = 4) = 5 P (X > ) = P (X ) = {P (X = ) + P (X = ) + P (X = )} = 3 = 7 P (.5 < X < 4.5/X > ) = 5 7 = 5 7 (iv) The smallest value of α for which P (X α) > =.5. Page 4 of 36

16 From the CDF table, we found that P (X 3) = 5 =.5. But we need the probability which is more than.5, for this we have P (X 4) = 8 condition. =.8 >.5. α = 4 satisfies the given Example: 7. A random variable X has the following distribution X P[X=x] k 3k 5k 7k 9k k 3k 5k 7k Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P ( < X < 3/X > ) and (iv) the smallest value of α for which P (X α) >. Hints/Solution: (i) The value of k = 8 X P [X = x] k 3k 5k 7k 9k k 3k 5k 7k P [X = x] (ii) the Distribution Function (CDF) X P [X = x] k 3k 5k 7k 9k k 3k 5k 7k P [X = x] F [X = x] = (iii) P ( < X < 3/X > ) P ( < X < 3/X > ) = P [( < X < 3) (X > )] P (X > ) P ( < X < 3) = P (X = ) + P (X = ) = 8 8 P (X > ) = P (X ) = {P (X = ) + P (X = ) + P (X = )} = 9 8 = 7 8 P (.5 < X < 4.5/X > ) = = 8 7 (iv) The smallest value of α for which P (X α) > =.5. Page 5 of 36

17 From the CDF table, we found that P (X 5) = 36 <.5. But we need the probability which is 8 more than.5, for this we have P (X 6) = 49 >.5. α = 6 satisfies the given condition. 8 kx, when x k, when x Example: 8. Find the value of k for the pdf f(x) =. Also find 3k kx, when x 3, otherwise (a) the CDF of X (b) P (.5 < X < 3./.5 < X <.8) Hints/Solution: i.e. i.e. f(x)dx + dx + [ x i.e. + k f(x)dx + kx dx + ] f(x)dx + k dx + We know that 3 3 [ + k [x] + 3k x k x x, when x (a) Since k =, f(x) =, when x (3 x), when x 3, otherwise f(x)dx + 3 (3k kx)dx + ] 3 f(x)dx = f(x)dx = 3 + = i.e. k = dx = = k = Page 6 of 36

18 The CDF of X is F (x) = P (X x) = f(x)dx x When x <, F (x) =, since f(x) = for x < x When x <, F (x) = f(x)dx When x <, F (x) = When x < 3, F (x) = = x = f(x)dx x dx + = x x x f(x)dx = f(x)dx x dx + dx + x x When x 3, F (x) = whenx < x, when x F (x) = (x ), when x 4 4 ( x + 6x 5), when x 3, when x 3 x x dx = x 4 dx = (x ) 4 (3 x) dx = 4 ( x + 6x 5) (b) P (.5 < X < 3./.5 < X <.8) = P [(.5 < X < 3.) (.5 < X <.8)] P (.5 < X <.8) Page 7 of 36

19 P [(.5 < X < 3.) (.5 < X <.8)] = P (.5 < X <.8).8 = dx = 3 P (.5 < X <.8) = P (.5 < X < 3./.5 < X <.8) =.5.5 = 6 x dx +.8 dx = 3 6 P [(.5 < X < 3.) (.5 < X <.8)] P (.5 < X <.8) Example: 9. Experience has shown that while walking in a certain park, the time X (in minutes) duration between seeing two people smoking has a density function { kxe x, when x > f(x) =, otherwise Find the (a) value of k (b) distribution function of X (c) What is the probability that a person, who has just seen a person smoking will see another person smoking in to 5 minutes, in at least 7 minutes? Hints/Solution: Given the pdf of X is { kxe x, when x > f(x) =, otherwise f(x) x = k >. (a) We know that f(x) dx = i.e. kxe x dx = i.e. k [x e x e x ( ) ] = = k = Page 8 of 36

20 { xe x, when x > f(x) =, otherwise (b) The distribution function (CDF) is given by F (x) = P (X x) = x f(x) dx x Here we have x >, F (x) = xe x dx = ( + x)e x, x > (c) P ( < X < 5) = = 5 5 f(x) dx xe x dx = 6e 5 + 3e = =.366 and P (X > 7) = = 7 f(x) dx 7 = 8e 7 =.7 xe x dx Page 9 of 36

21 Example:. The sales of convenience store on a randomly selected day are X thousand dollars, where X is a random variable with a distribution function, when x < x F (x) =, when x < k(4x x ), when x <, when x Suppose that this convenience store s total sales on any given day is less than units(dollars or Pounds or Rupees). Find the (a) value of k (b) Let A and B be the events that tomorrow the store s total sales between 5 and 5 units respectively. Find P (A) and P (B). (c) Are A and B are independent events? Hints/Solution: The pdf of X is given by, when x < f(x) = F (x) = d dx F (x) = x, when x < k(4 x), when x <, when x Since f(x) is a pdf, i.e. i.e. f(x)dx + dx + f(x)dx + x dx + we have f(x)dx + [ x i.e. + ] k(4 x) dx + f(x)dx = f(x)dx = + k [ 4x x ] + = dx = = k =, when x < x, when x < f(x) = ( x), when x <, when x Page of 36

22 Since the total sales X is in thousands of units, the sales between 5 and 5 units is the event A which stands for =.5 < X < 3 =.5 and the sales over units is the event B which stands for X >. = A B = < X <.5 Now P (A) = P (.5 < X <.5) =.5 f(x)dx = x dx + ( x) dx = 3 4 P (B) = P (X > ) = = ( x) dx = f(x)dx P (A B) = P ( < X <.5) = =.5 ( x) dx = 3 8 The condition for independent events: P (A) P (B) = P (A B) Here, P (A) P (B) = 3 4 = 3 = P (A B) 8 A and B are independent events..5 f(x)dx Example:. If a random variable X has the following probability distribution, find E(X), E(X x ), V ar(x), E(X + ), V ar(x + ). p(x) Hints/Solution: Here X is a discrete RV. E(X) = x i p(x i ) i= = ( ) = =.5 Page of 36

23 E(X ) = x i p(x i ) i= = ( ) = =.5 V ar(x) = E(X ) [E(X)] = (.5) (.5) =.5.5 =.5 E(X + ) = E(X) + = (.5) + = + = V ar(x + ) = V ar(x) = 4 (.5) = 5 Example:. If a random variable X has the following probability distribution, find x 3 4 E(X), E(X ), V ar(x), E(3X 4), V ar(3x 4) p(x) Hints/Solution: Here X is a discrete RV. E(X) = i= = 4 5 x i p(x i ) E(X ) = i= = 46 5 x i p(x i ) Page of 36

24 V ar(x) = E(X ) [E(X)] = 46 ( ) = 34 5 E(3X 4) = 3E(X) 4 = = 5 V ar(3x 4) = 3 V ar(x) = 36 5 Example: 3. A test engineer found that the distribution function of the lifetime X (in years) of an equipment follows is given by {, when x < F (x) = e x 5, when x Find the pdf, mean and variance of X. Hints/Solution: The pdf of X is given by f(x) = F (x) = d dx F =, when x < x 5, when x 5 e The Mean E(X) = xf(x)dx = x 5 e x 5 dx = 5 [ ] (5x + 5)e x 5 = 5 V ar(x) = E(X ) [(X)] Page 3 of 36

25 Now E(X ) = x f(x)dx = x 5 e x 5 dx = 5 [ ] (5x + 5x + 5)e x 5 = 5 V ar(x) = 5 [5] = 5 Example: 4. Find the mean and standard deviation of the distribution { kx( x), when < x f(x) =, otherwise Hints/Solution: Given that the continuous RV X whose pdf is given by { kx( x), when < x f(x) =, otherwise Since f(x) is a pdf, i.e. f(x)dx + we have f(x)dx + f(x)dx = f(x)dx = i.e. dx + kx( x) dx + dx = i.e. + k ] [x x3 + = 3 = k = 3 4 Page 4 of 36

26 The Mean E(X) = xf(x)dx = x 3 x( x)dx 4 = 3 4 ] [ x3 3 x4 = 4 V ar(x) = E(X ) [(X)] Now E(X ) = = = 3 4 V ar(x) = 6 5 [] = 5 x f(x)dx x 3 x( x)dx 4 [ x4 4 x5 5 Example: 5. If X has the distribution function, when x < ] = 6 5 = S.D. = 5 3, when x < 4 F (x) =, when 4 x < 6 5, when 6 x < 6, when x Find (a) the probability distribution of X (b) Find the mean and variance of X. Page 5 of 36

27 Hints/Solution: Given that the CDF of X is, when x < 3, when x < 4 F (x) =, when 4 x < 6 5, when 6 x < 6, when x We know that P (X = x i ) = F (x i ) F (x i ), x i x x i. i =,, 3,..., where F is constant in The CDF changes its values at x =, 4, 6,. The probability distribution takes its values as follows: The probability distribution is given by P (X = ) = F () F () = 3 = 3 P (X = 4) = F (4) F () = 3 = 6 P (X = 6) = F (6) F (4) = 5 6 = 3 P (X = ) = F () F (6) = 5 6 = 6 Mean=E(X) = Σxp(x) = 4 3. E(X ) = Σx p(x) = V ar(x) = E(X ) [E(X)] = = x 4 6 p(x) Page 6 of 36

28 Example: 6. If X has the distribution function, when x < 6, when x < F (x) =, when x < 4 5 8, when 4 x < 6, when x 6 Find (a) the probability distribution of X (b) Find the mean and variance of X. Hints/Solution: Given that the CDF of X is, when x < 6, when x < F (x) =, when x < 4 5 8, when 4 x < 6, when x 6 We know that P (X = x i ) = F (x i ) F (x i ), x i x x i. i =,, 3,..., where F is constant in The CDF changes its values at x =, 4, 6,. The probability distribution takes its values as follows: P (X = ) = 6 P (X = ) = F () F () = 6 = 3 P (X = 4) = F (4) F () = 5 8 = 8 P (X = 6) = F (6) F (4) = 5 8 = 3 8 The probability distribution is given by x 4 6 p(x) Page 7 of 36

29 Mean=E(X) = Σxp(x) = 37. E(X ) = Σx p(x) = 6. V ar(x) = E(X ) [E(X)] = 6 Probability & Statistics [ ] 37 = = 7.3. Example: 7. The first four moments about X = 5 are,5, and 48. Find the first four central moment. Hints/Solution: Let m, m, m 3, m 4 be the first four moments about X = 5. Given m =, m = 5, m 3 =, m 4 = 48 be the first four moments about X = 5 m = E(X 5) = = E(X) 5 = = E(X) = µ = 6 µ = m (m ) = 5 = 4 µ 3 = m 3 3m m + (m ) 3 = () = µ 4 = m 4 4m 3 m + 6m (m ) 3(m ) 4 = 48 4()() + 6(5)() 3() = 7 Example: 8. The first four moments about x = 4 are,4, and 45. Find the first four moments about the mean. Hints/Solution: Let m, m, m 3, m 4 be the first four moments about X = 4. Given m =, m = 4, m 3 =, m 4 = 45 be the first four moments about X = 4 m = E(X 4) = = E(X) 4 = = E(X) = µ = 5 µ = m (m ) = 4 = 3 µ 3 = m 3 3m m + (m ) 3 = () = µ 4 = m 4 4m 3 m + 6m (m ) 3(m ) 4 = 45 4()() + 6(4)() 3() = 6 Example: 9. A random variable X has the pdf f(x) = kx e x, x. Find the r th moment and hence find the first four moments. Page 8 of 36

30 Hints/Solution: We know that f(x) dx = Probability & Statistics i.e., kx e x dx = i.e., k e x x 3 dx = i.e., kγ(3) = Now, The r th moment is given by Γ(n) = e x x n dx i.e., k! = = k =. [ Γ(n) = (n )!] µ r = E(X r ) = = = k x r f(x) dx x r kx e x dx e x x r+3 dx = Γ(r + 3) = (r + )! Now, First Moment µ = (3)! = 3 Second Moment µ = (4)! = Third Moment µ 3 = (5)! = 6 Fourth Moment µ 4 = (6)! = 36 Example:. A random variable X has the pdf f(x) = e x, x. Find the MGF(Moment Generating Function) and hence find its mean and variance. Page 9 of 36

31 Hints/Solution: The MGF of X is given by M X (t) = E(e tx ) = e tx f(x) dx = e tx e x dx M X (t) = Now, Differentiating w.r.to t, we get = [ = e ( t)x dx e ( t)x ( t) ] t if t <. = ( t) = + t + 4t + 8t M X(t) = + 8t + 4t +... M X(t) = t +... Now, First Moment=Mean= E(X) = µ = M X() = Second Moment= E(X ) = µ = M X() = 8 V ar(x) = E(X ) [E(X)] = 8 4 = 4. Example:. A random variable X has the pdf p(x) =, x =,, 3,.... Find the x MGF(Moment Generating Function) and hence find its mean and variance. Page 3 of 36

32 Hints/Solution: The MGF of X is given by M X (t) = E(e tx ) = M X (t) = Now, Differentiating w.r.to t, we get x= e tx p(x) = e tx = ( ) e t x ( e t = x x= x= x= [ ( ) ( ) ] = et e t e t [ ( )] = et e t e t e t if e t. M X(t) = M e t ( e t ) X(t) = [( et )e t + e t ] ( e t ) 3 Now, First Moment=Mean= E(X) = µ = M X() = Second Moment= E(X ) = µ = M X() = 6 V ar(x) = E(X ) [E(X)] = 6 4 =. Example:. A random variable X has the r th moment of the form µ r = (r + )! r. Find the MGF(Moment Generating Function) and hence find its mean and variance. ) x Page 3 of 36

33 Hints/Solution: The MGF of X is given by Given µ r = (r + )! r µ =! µ = 3! µ 3 = 4! 3. M X (t) = E(e tx ) = + t! µ + t! µ + t3 3! µ tr r! µ r +... = + t! + t! 3! + t3 3! 4! = + (t) + 3(t) + 4(t) M X (t) = ( t). Now, Differentiating w.r.to t, we get M X(t) = ( t) 3 ( ) M X(t) = 6( t) 4 ( ) Now, First Moment=Mean= E(X) = µ = M X() = 4 Second Moment= E(X ) = µ = M X() = 4 V ar(x) = E(X ) [E(X)] = 4 6 = 8. Page 3 of 36

34 6 Exercise/Practice/Assignment Problems. Find the probability of drawing two red balls in succession from a bag containing 3 red and 6 black balls when (i) the ball that is drawn first is replaced, (ii) it is not replaced.. A bag contains 3 red and 4 white balls. Two draws are made without replacement; what is the probability that both the balls are red. 3. Box contains bulbs of which % are defective. Box contains bulbs of which 5% are defective. Two bulbs are drawn (without replacement) from a randomly selected box. (i) Find the probability that both balls are defective and (ii) assuming that both are defective, find the probability that they came from box. 4. The chance of a doctor D will diagnose a disease Z correctly is 6%. The chance that the patient will survive by his treatment after wrong diagnosis is 4% and the chance of survive for correct diagnosis is 7%. Find the probability of the patient survival. A patient of doctor D, who has disease Z, survives. What is the chance that his disease was diagnosed wrongly? 5. A random variable X has the following distribution X P[X=x]. k. k.3 3k Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P ( < X < 3/X < ) and (iv) the smallest value of α for which P (X α) >. Ans: k = 5 6. A random variable X has the following distribution X 3 4 P[X=x] k k 5k 7k 9k Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P ( < X < 3/X < ) and (iv) the smallest value of α for which P (X α) > 3. Ans: k = 4 7. A random variable X has a pdf { 3x, when x f(x) =, otherwise Find the (i) Find a, b such that P (X a) = P (X > a) and P (X > b) =.5,(ii) the Distribution Function (CDF) (iii) P ( < X <.5/X <.8). Ans: a = 3, b = The amount of bread (in hundred kgs) that a certain bakery is to sell in a day is a random variable X with a pdf Ax, when x < 5 f(x) = A( x), when 5 x <, otherwise Page 33 of 36

35 Find the (i) the value of A,(ii) the Distribution Function (CDF) (iii) P (X > 5/X < 5), P (X > 5/.5 < X < 7.5). (iv) the probability that in a day the sales is (a) more than 5 kgs (b) less than 5 kgs (c) between 5 and 75 kgs. Ans: A = 5 9. The cumulative distribution function (CDF) of a random variable X is given by 4 F (x) = x, when x >, otherwise Find the (i) the pdf of X,(ii) P (X > 5/X < 5), P (X > 5/.5 < X < 7.5) (iii) P (X < 3), P (3 < X < 5).. A coin is tossed until a head appears. What is the expected value of the number of tosses?. Also find its variance.. The pdf of a random variable X is given by { a + bx, when x f(x) =, otherwise Find the (i) the value of a, b if the mean is /,(ii) the variance of X (iii) P (X >.5/X <.5). The first three moments about the origin are 5,6,78. Find the first three moments about the value x=3. Ans:,5, The first two moments about x=3 are and 8. Find the mean and variance. Ans: 4,7 4. The pdf of a random variable X is given by { k( x), when x f(x) =, otherwise Find the (i) the the value of k,(ii) the r th moment about origin (iii) mean and variance. Ans: k = 5. An unbiased coin is tossed three times. If X denotes the number of heads appear, find the MGF of X and hence find the mean and variance. 6. Find the MGF of the distribution whose pdf is f(x) = ke x, x > and hence find its mean and variance. 7. The pdf of a random variable X is given by x, when x f(x) = x, when < x, otherwise For this find the MGF and prove that mean and variance cannot be find using this MGF and then find its mean and variance using expectation. Page 34 of 36

36 8. The pdf of a random variable X is given by f(x) = ke x, < x < Find the (i) the the value of k,(ii) the r th moment about origin (iii) the MGF and hence mean and variance. Ans: k = 9. Find the MGF of the RV whose moments are given by µ r = (r)!. Find also its mean and variance. SOLVE MORE PROBLEMS FROM THE TEXTBOOK AND REFERENCE BOOKS Figure : Values of e λ. Contact: (+9) (or) athithan.s@ktr.srmuniv.ac.in Visit: Page 35 of 36

37 LE C TU R E N O TE S O F AT H IT H A N S Probability & Statistics Figure : Area under Standard Normal Curve. Page 36 of 36