Question

Problem 6 a. Let X and Y be independent identically distributed exponential random variables with probability density function fX(x) = ?e^-?x and fY(y) = ?e^-?y respectively. Define W = X / min(X, 2Y) Find the p.d.f of W. b. Let fXY(x, y) = { 2e^-(x+y) 0 < x < y < ?; 0 otherwise } Define Z = X + Y W = Y/X Determine the joint pdf of Z and W. Are Z and W independent random variables?

          Problem 6

a. Let X and Y be independent identically distributed exponential random variables with probability density function fX(x) = ?e^-?x and fY(y) = ?e^-?y respectively. Define

W = X / min(X, 2Y)

Find the p.d.f of W.

b. Let

fXY(x, y) = { 2e^-(x+y) 0 < x < y < ?; 0 otherwise }

Define

Z = X + Y W = Y/X

Determine the joint pdf of Z and W. Are Z and W independent random variables?

Problem 6

a. Let X and Y be independent identically distributed exponential random variables with probability density function fX(x) = ?e^-?x and fY(y) = ?e^-?y respectively. Define

W = X / min(X, 2Y)

Find the p.d.f of W.

b. Let

fXY(x, y) = 2e^-(x+y) 0 < x < y < ?; 0 otherwise

Define

Z = X + Y W = Y/X

Determine the joint pdf of Z and W. Are Z and W independent random variables?

Added by Jorge C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Adi S

09/04/2023

Step 1

We have: $F_W(w) = P(W \le w) = P(\min(X, 2Y) \le w)$ Since X and Y are independent, we can write this as: $F_W(w) = P(X \le w, 2Y \le w) = P(X \le w)P(2Y \le w)$ Now, we find the CDF of X and Y: $F_X(x) = P(X \le x) = 1 - e^{-\lambda x}$ $F_Y(y) = P(Y \le Show more…

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Problem 6 a. Let X and Y be independent identically distributed exponential random variables with probability density function fX(x) = λe^-λx and fY(y) = λe^-λy respectively. Define W = X / min(X, 2Y) Find the p.d.f of W. b. Let fXY(x, y) = { 2e^-(x+y) 0 < x < y < ∞; 0 otherwise } Define Z = X + Y W = Y/X Determine the joint pdf of Z and W. Are Z and W independent random variables?