Solution to Assignment 3

Transcription

1 The Chinese University of Hong Kong ENGG3D: Probability and Statistics for Engineers 5-6 Term Solution to Assignment 3 Hongyang Li, Francis Due: 3:pm, March Release Date: March 8, 6 Dear students, The solution is now released. In this assignment, most of the problems deal with the calculation of PDF, integration, expectation, etc. A brief summary is listed as follows:. Problem., the conditional PDF given an even is introduced in Section 3.5 in the textbook, it s a density function of x, instead of a number.. I find the calculation of integration to be the gotcha part. Many students don t know to apply integration by parts in some cases (P3.3, P.3). We assume you are capable of doing some complicated integrations, if not, see this online tutorial about the integration by parts. 3. In Problem, the case where a < is not considered. We can t just assume x falls into [, a] without any constraint on a.. The range of the integration (P3.) or random (P6) variable is not taken care of. You need to give a complete equation form spanning all the cases of the variables. See the solution for details. I strongly encourage you to review back of your assignment and see what s going on with your own solutions. My office hour for assignment 3 is 7: - 9: pm, Monday, March, SHB 3. Finally, this document can also be accessed externally through this link.

2 Problem (8 points). A continuous random variable X s PDF is defined as { a/x x [, ], otherwise. where a is a constant.. Find a.. Find mean and variance of X. 3. Find CDF of X.. The integration of one variable s PDF has to be, therefore: [ a = x dx = a ] x. The expectation is First we compute E[X] = Therefore, the variance is 3. For x, we have E[X ] = a = ( ) [ ] x x dx = ln x = ln =.77 x ( x )dx = [ ] x = 8 V ar[x] = E[X ] (E[X]) = 8 ( ln ) =.3 F X (x) = x t dt = x For x, F X (x) = and for x, F X (x) =. Problem (8 points). A continuous random variable X s PDF is defined as { x/ x [, 3], otherwise. and A is the event {X }.

3 . Calculate E[X] and P (A).. Find the conditional PDF of X given that A has occurred. 3. Let Y = X. Find E[Y ] and V ar(y ).. E[X] = 3 ( ) x x dx = 3 6 P (A) = 3 ( ) x dx = 5 8. From Section 3.5 in the textbook, we have { fx (x) P (A) f X X A (x) = = x 5 if x [, 3] otherwise 3. E[Y ] = E[X ] = 3 x ( x ) dx = 5 V ar(y ) = V ar(x ) = E[X ] ( E[X ] ) = = 6 3 Problem 3 (8 points). Two continuous random variables are uniformly distributed in the following region, {(x, y) x + y r, y } for some given constant number r.. Find the joint PDF of X and Y.. Find the marginal PDF of X and Y. 3. Find E[X] and V ar(x).. The joint distribution is obtained by over the area of the semi-circle f X,Y (x, y) = πr, {(x, y) x + y r, y } 3

4 . f Y (y) = r y f(x, y)dx = r y πr dx = r πr y, y [ r, ] f(x, y)dy = r x πr dy = πr r x, x [ r, r] Remark: Pay attention to the range of y, which is below zero, and the integration bound when we compute the marginal PDF. 3. E[X] = r r x r πr x dx = [ πr ] r 3 (r x ) 3 = r To get the variance, first we compute E[X ] = r r x πr r x dx Let t = x and thus dx = dx, we have E[X ] = r πr t r tdt Applying the rule of integration by parts, let u = t and v = r t, we have Then the variance of X is E[X ] = [ πr ] 3 t(r t) 3 r 5 (r t) 5 = r3 5π V ar(x) = E[X ] (E[X]) = r3 5π Remark: This question is not for the faint of heart. You should have a good ability of computing (a little bit) complicated integration. Problem (8 points). An exponential random variable s PDF is defined as { λe λx x,, otherwise. Let A be the event of {x a} with some constant scalar value a. uv dt = uv vu dt, where both u and v are functions of t.

5 . Find P (A).. Find f X A (x). (Note: There is a typo in the assignment. p should be f) 3. Find E[X A].. For a, P (A) =. For a >, we have P (A) = a λe λx dx = [ e λx] a = e λa. Remark: Your score will be deducted by 3 points if a is not considered.. For a, f X A (x) =. For a >, we have { f X A (x) = f λe λx X(x) P (A) = if x a, e λa otherwise. 3. For a, E[X A] =. For a >, by applying integration by parts, we have E[X A] = = a xf X A (x)dx = λ e λa a [ ( x λ + λ x λe λx dx e λa ) ] a e λx = e λa λae λa ( e λa )λ Problem 5 (8 points). There is a machine manufacturing some product. The length of the product is a normal random variable with a mean of cm and a standard deviation of cm.. What is the probability that a product s length is within the interval of [, ].. If we have another machine that has the same specification as the first one, where two machines operate independently and simultaneously. What is the probability that both products are within the interval of [, ].. Let X be the length of the product and Z = X µ σ, we have µ =, σ =, ) ( P ( X ) = P Z. P = P A ( X ) P B ( X ) = (.686) =.66 = P ( Z ) =.686 5

6 Problem 6 ( points). If X is a continuous random variable uniform on [, ].. Find the PDF of X.. Find the PDF of ln X. The PDF of X is { 9 if x, otherwise.. Let Y = g(x) = X, since g is strictly monotonic and its inverse is h(y) = y, we have the PDF of Y as follows f X (h(y)) f Y (y) = dh(y) dy = 9 y if y, otherwise.. Let Z = g(x) = ln X, since g is strictly monotonic and its inverse is h(z) = e z, we have the PDF of Z as follows f X (h(z)) f Z (z) = dh(z) dz = 9 e z if ln z, otherwise. 6