Question

Suppose Y is a continuous random variable that has a pdf given by f(y) = c exp{-y - e^{-y}} for y ? ? for some constant c, and cdf F(y) = ?_{-?}^{y} c exp{-t - e^{-t}}dt = c exp{-e^{-y}} for y ? ?. (a) Write down the value of c. (b) Compute P(Y > 1). (c) Find the pdf of random variable X = e^{-Y}, and hence identify the distribution of X. Hint: For x > 0, P(X ? x) = P(e^{-Y} ? x) = P(Y ? -ln x).

          Suppose Y is a continuous random variable that has a pdf given by

f(y) = c exp{-y - e^{-y}}

for y ? ? for some constant c, and cdf

F(y) = ?_{-?}^{y} c exp{-t - e^{-t}}dt = c exp{-e^{-y}}

for y ? ?.

(a) Write down the value of c.
(b) Compute P(Y > 1).
(c) Find the pdf of random variable X = e^{-Y}, and hence identify the distribution of X.
Hint: For x > 0,

P(X ? x) = P(e^{-Y} ? x) = P(Y ? -ln x).

Suppose Y is a continuous random variable that has a pdf given by

f(y) = c exp-y - e^-y

for y ? ? for some constant c, and cdf

F(y) = ?-?^y c exp-t - e^-tdt = c exp-e^-y

for y ? ?.

(a) Write down the value of c.
(b) Compute P(Y > 1).
(c) Find the pdf of random variable X = e^-Y, and hence identify the distribution of X.
Hint: For x > 0,

P(X ? x) = P(e^-Y ? x) = P(Y ? -ln x).

Added by Jane S.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Madhur L

08/23/2023

Step 1

To do this, we know that the integral of the pdf over its entire domain must equal 1. So, we have: $$ \int_{-\infty}^{\infty} c \exp{-y - e^{-y}} dy = 1 $$ Now, we can use the given cdf to help us find the integral: $$ F(y) = \int_{-\infty}^{y} c \exp{-t - Show more…

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Suppose Y is a continuous random variable that has a pdf given by f(y) = c exp{-y - e^-y} for y ∈ ℝ for some constant c, and cdf F(y) = ∫_{-∞}^{y} c exp{-t - e^-t}dt = c exp{-e^-y} for y ∈ ℝ. (a) Write down the value of c. (b) Compute P(Y > 1). (c) Find the pdf of random variable X = e^-Y, and hence identify the distribution of X. Hint: For x > 0, P(X ≤ x) = P(e^-Y ≤ x) = P(Y ≥ -ln x).