Find the pdf $f_W(w)$ if the joint pdf of $X,Y>0$ is $f(x,y)$ and $W=Y/X$

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Exercise :

Let $X>0$ and $Y>0$ continuous random variables with joint probability density function (pdf) $f(x,y)$. If $W=Y/X$, then find the pdf $f_W(w)$. What happens to $f_W(w)$ and its expression when $X,Y$ are independent ?

Attempt-Question :

$$F_W(w)=P\{W \leq w\} = P\{Y/X \leq w\}=\int \int_Df(x,y)dxdy$$

$$\Rightarrow$$

$$F_W(w) = \int\bigg[\int f(x,y)dx\bigg]dy$$

It is : $Y/X \leq w \Leftrightarrow Y \leq wX$ since $X>0$. This means that on a $X-Y$ axis system, we'll need every value below $wX$.

$$=\int_{-\infty}^{ωx}\bigg[\int_{-\infty}^{+\infty}f(x,y)dx\bigg]dy$$

Which leads to :

$$f_W(w) = \frac{dF_W(w)}{dw} = \frac{d}{dw}\int_{-\infty}^{ωx}\bigg[\int_{-\infty}^{+\infty}f(x,y)dx\bigg]dy = \int_{-\infty}^\infty\bigg[\frac{d}{dw}\int_{-\infty}^{wx}f(x,y)dy\bigg]dx$$

Question : Is my approach and the integral bounds correct ? How would I take it one step further from then on ?