Question

Let the continuous random variables X and Y have joint pdf f(x,y)=e^{-x-y}, 0<x<\infty, 0<y<\infty, then f(\frac{y}{x})=a)e^{-y}b)e^{-x}c)\frac{e^{y}}{e^{-x}}

Random variables
ANSWERED
asked 2021-05-13

Let the continuous random variables X and Y have joint pdf

\(f(x,y)=e^{-x-y}, 0\)

a)\(e^{-y}\)

b)\(e^{-x}\)

c)\(\frac{e^{y}}{e^{-x}}\)

Answers (1)

2021-05-14

Step 1
The joint pdf of the continuous random variables X and Y is given by :
\(f(x,y)=e^{-x-y}, 0\)
Here we have to find the value of \(f(\frac{y}{x})\).
As we know that \(f(\frac{y}{x})=\frac{f(x,y)}{f_{X}(x)}\)
And \(f_{X}(x)=\int_{a}^{b}f(x,y)dy\)
Step 2
Let us finding the required value:
\(f(\frac{y}{x})=\frac{f(x,y)}{\int_{a}^{b}f(x,y)dy}\)
\(=\frac{e^{-x-y}}{\int_{0}^{\infty}e^{-x-y}dy}\)
\(=\frac{e^{-x-y}}{[-e^{-x-y}]_{0}^{\infty}}\)
\(=\frac{e^{-x-y}}{[-e^{-\infty}e^{-x}]}\)
\(=\frac{e^{-x-y}}{e^{-x}}\)
\(=e^{-x-y+x}\)
\(=e^{-y}\)
Thus, the required value of \(f(\frac{y}{x})=e^{-y}\).
Hence, option (a) is correct.

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