Continuous distributions

Transcription

1 CHAPTER 7 Continuous distributions 7.. Introduction A r.v. X is said to have a continuous distribution if there exists a nonnegative function f such that P(a X b) = ˆ b a f(x)dx for every a and b. distribution.) f is called the density function for X. Note ) =. In particular, P(X = a) = a a (More precisely, such an X is said to have an absolutely continuous f(x)dx = P( < X < f(x)dx = for every a. Example 7.: we have c =. Suppose we are given c/x for x. Since c f(x)dx = c x dx = c, f(x)dx = and Define F (y) = P( < X y) = y f(x)dx. F is called the distribution function of X. We can define F for any random variable, not just continuous ones, by setting F (y) = P(X y). In the case of discrete random variables, this is not particularly useful, although it does serve to unify discrete and continuous random variables. In the continuous case, the fundamental theorem of calculus tells us, provided f satisfies some conditions, that f(y) = F (y). By analogy with the discrete case, we define the expectation by In the example above, E X = E X = xf(x)dx. x x dx = x dx =. We give another definition of the expectation in the continuous case. First suppose X is nonnegative. Define X n (ω) to be k/ n if k/ n X(ω) < (k + )/ n. We are approximating X from below by the largest multiple of n. Each X n is discrete and the X n increase to X. We define E X = lim n E X n. 8

2 8 7. CONTINUOUS DISTRIBUTIONS Let us argue that this agrees with the first definition in this case. We have E X n = k P(X n n = k/ n ) k/ n = k n P(k/n X < (k + )/ n ) k/ n = ˆ k (k+)/ n f(x)dx n k/ n = ˆ (k+)/ n k/ n k n f(x)dx. If x [k/ n, (k +)/ n ), then x differs from k/ n by at most / n. So the last integral differs from ˆ (k+)/ n xf(x)dx k/ n by at most (/ n )P(k/ n X < (k + )/ n ) / n, which goes to as n. On the other hand, ˆ (k+)/ ˆ n M xf(x)dx = xf(x)dx, k/ n which is how we defined the expectation of X. We will not prove the following, but it is an interesting exercise: if X m is any sequence of discrete random variables that increase up to X, then lim m E X m will have the same value E X. To show linearity, if X and Y are bounded positive random variables, then take X m discrete increasing up to X and Y m discrete increasing up to Y. Then X m + Y m is discrete and increases up to X + Y, so we have E (X + Y ) = lim E (X m + Y m ) m = lim E X m + lim E Y m = E X + E Y. m m If X is not necessarily positive, we have a similar definition; we will not do the details. This second definition of expectation is mostly useful for theoretical purposes and much less so for calculations. Similarly to the discrete case, we have Proposition 7.: E g(x) = g(x)f(x)dx. As in the discrete case, VarX = E [X E X]. As an example of these calculations, let us look at the uniform distribution. We say that a random variable X has a uniform distribution on [a, b] if f X (x) = if a x b and b a otherwise.

3 To calculate the expectation of X, E X = 7.. INTRODUCTION 8 xf X (x)dx = ˆ b a x b a dx = x dx b a a = ( b ) b a a = a + b. This is what one would expect. To calculate the variance, we first calculate E X = We then do some algebra to obtain ˆ b x f X (x)dx = ˆ b VarX = E X (E X) = a x b a dx = a + ab + b. (b a).

4 84 7. CONTINUOUS DISTRIBUTIONS Example 7.: 7.. Further examples and applications Suppose X has the following p.d.f. x x x. Find the c.d.f of X, that is, find F X (x). Use this c.d.f to find P ( X 4). Answer. We have F X (x) = if x and will need to compute F X (x) = P (X x) = y dy = x when x. We can use this formula to find the following probability: P ( X 4) = P (X 4) P (X < ) = F X (4) F X () = ( 4 ) Example 7.: Find EX. Answer. We have that Example 7.4: Find E [ e X]. Suppose X has density ˆ E [X] = ˆ x x x otherwise. xf(x)dx = ˆ x x dx =. The density of X is given by if x. otherwise Answer. Using proposition 7. with g(x) = e x we have Example 7.5: Find Var(X). Ee X = ˆ Suppose X has density e x dx = [ e ]. x x otherwise. Answer. From Example 7. we found E [X] =. Now E [ X ] = ˆ x xdx = Copyright 7 Phanuel Mariano, Patricia Alonso-Ruiz. ˆ x dx =. ( ) = 7 44.

5 7.. FURTHER EXAMPLES AND APPLICATIONS 85 Thus Var(X) = ( ) = 8. Example 7.6: Suppose X has density ax + b x otherwise. and that E [X ] =. Find the values of a and b. 6 Answer. We need to use the fact that f(x)dx = and E [X ] =. The first one gives 6 us, and the second one give us = 6 = Solving these equations gives us ˆ ˆ (ax + b) dx = a + b x (ax + b) dx = a 4 + b. a =, and b =.

6 86 7. CONTINUOUS DISTRIBUTIONS 7.. Exercises Exercise 7.: Let X be a random variable with probability density function cx (5 x) x 5, otherwise. (a) What is the value of c? (b) What is the cumulative distribution function of X? That is, find F X (x) = P (X x). (c) Use your answer in part (b) to find P ( X ). (d) What is E [X]? (e) What is Var(X)? Exercise 7.: UConn students have designed the new U-phone. They have determined that the lifetime of a U-Phone is given by the random variable X (measured in hours), with probability density function x, x x. (a) Find the probability that the u-phone will last more than hours. (b) What is the cumulative distribution function of X? That is, find F X (x) = P (X x). (c) Use part (b) to help you find P (X 5)? Exercise 7.: Suppose the random variable X has a density function x x >, x. Compute E [X]. Exercise 7.4: An insurance company insures a large number of homes. The insured value, X, of a randomly selected home is assumed to follow a distribution with density function x >, x 4 otherwise. Given that a randomly selected home is insured for at least.5, calculate the probability that it is insured for less than. Exercise 7.5: The density function of X is given by a + bx x, otherwise. If E [X] = 7, find the values of a and b.

7 Exercise 7.6: 7.. EXERCISES 87 Let X be a random variable with density function < x < a, a otherwise. Suppose that E [X] = 6Var(X). Find the value of a. Exercise 7.7: Suppose you order a pizza from your favorite pizzaria at 7: pm, knowing that the time it takes for your pizza to be ready is uniformly distributed between 7: pm and 7: pm. (a) What is the probability that you will have to wait longer than minutes for your pizza? (b) If at 7:5pm, the pizza has not yet arrived, what is the probability that you will have to wait at least an additional minutes? Exercise 7.8: The grade of deterioration X of a machine part has a continuous distribution on the interval (, ) with probability density function f X (x), where f X (x) is proportional to x on the interval. The reparation costs of this part are modeled by a random variable Y 5 that is given by Y = X. Compute the expected cost of reparation of the machine part. Exercise 7.9: You arrive at a bus stop at : 5 am, and the bus arrives at some (random) time uniformly distributed between : and :. Given that when you arrive today to the station the bus is not there yet (you are lucky today), what is the probability that you have to wait more than 4 minutes? Hint: the event today you are lucky can be expressed as (X > 5), where X denotes the arrival time of the bus at the station (in minutes past : a.m.).

8 88 7. CONTINUOUS DISTRIBUTIONS 7.4. Selected solutions Solution to Exercise 7.(A): We must have that f(x)dx =, thus ˆ 5 [ ( )] 5x 5 = cx(5 x)dx = c x and so we must have that c = 6/5. Solution to Exercise 7.(B): We have that ˆ x F X (x) = P (X x) = f(y)dy ˆ x [( 6 6 5y = y (5 y) dx = 5 5 y = 6 ( ) 5x 5 x. )] x Solution to Exercise 7.(C): We have P ( X ) = P (X ) P (X < ) = 6 ( ) 5 6 ( =.96. ) Solution to Exercise 7.(D): We have E [X] = =.5. xf X (x)dx = ˆ 5 Solution to Exercise 7.(E): We need to first compute Then E [ X ] = = 7.5. x f X (x)dx = ˆ 5 x x 6 x(5 x)dx 5 6 x(5 x)dx 5 Var(X) = E [ X ] (E [X]) = 7.5 (.5) =.5. Solution to Exercise 7.(A): We have x dx =. Solution to Exercise 7.(B): We have F (x) = P(X x) = for x >, and F (x) = for x <. ˆ x dy = y x

9 Solution to Exercise 7.(C): We have Solution to Exercise 7.5: The first one gives us, and the second one give us 7.4. SELECTED SOLUTIONS 89 P (X 5) = P (X < 5) = F X (5) ( = ) = 5 5. We need to use the fact that 7 f(x)dx = and E [X] =. = 7 = Solving these equations gives us Solution to Exercise 7.6: Also then we need Then ˆ ˆ Note that EX = EX = V ar(x) = ( a + bx ) dx = a + b x ( a + bx ) dx = a + b 4. a =, and b = 5 5. ˆ a x a dx = a +. Var(X) = EX (EX) ˆ a x a dx = a + a +. ( a + a + ) ( a + ) = a 6 a +. Then, using E [X] = 6Var(X), we simplify and get a a =, which gives us a =. Another way to solve this problem is to note that, for the uniform distribution on [a, b], the mean is a+b and the variance is (a b). This gives us an equation 6 (a ) = a+. Hence (a ) = a +, which implies a =. Solution to Exercise 7.7(A): Note that X is uniformly distributed over (, ). Then P(X > ) =. Solution to Exercise 7.7(B): Note that X is uniformly distributed over (, ). Then P(X > 5 X > 5) = P (X > 5) P(X > 5) = 5/ 5/ = /.

10 9 7. CONTINUOUS DISTRIBUTIONS Solution to Exercise 7.8: First of all we need to find the pdf of X. So far we know that cx x, 5 otherwise. Since c x dx = c, we have c =. Now, applying Proposition 7., 5 ˆ E[Y ] = 5 x dx = 5.