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.