snowlovelydayM

2021-11-06

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

jlo2niT

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

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