# 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

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)$$\frac{e^{y}}{e^{-x}}$$

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.