# Thread: pdf of function of two random variables (It is little difficult)

1. ## cdf of function of two random variables (It is little difficult)

Hello
I am Ph.D. student studying electrical engineering.
I have a question related with cdf of function of two random variables.
The question is attached in this post as pdf file.

2. Hello,

(a) is ... weird Oo

First thing : if X and Y are independent, you can write the product $f_X(x')f_Y(y')$
if not, you have to find the joint pdf of X and Y.

Second thing :
The whole formula is not very correct...

Assuming they're independent :

\begin{aligned}
P\left(\tfrac{X}{1+Y}\leq z\right)=P(X \leq z(1+Y))
&=P\left(Y\geq \tfrac{X-z}{z}\right) \\
&=\int_{x'=0}^\infty \int_{y'=0}^{\tfrac{x'-z}{z}} f_Y(y')f_X(x') ~dy' ~dx' \quad (*) \\
&=\int_{x'=0}^\infty f_X(x')\left(\int_{y'=0}^{\tfrac{x'-z}{z}} f_Y(y') ~dy'\right) ~dx' \\
&=\int_{x'=0}^\infty f_Y(y') F_Y\left(\tfrac{x'-z}{z}\right) ~dx \\
&=\mathbb{E}\left[F_Y\left(\tfrac{X-z}{z}\right)\right]
\end{aligned}

For the boundaries in (*), that's using the fact that $P\left(Y\geq \frac{X-z}{z}\right)=\mathbb{E}\left[\bold{1}_{Y\geq \frac{X-z}{z}}\right]$

and that for any measurable function h, $\mathbb{E}[h(X,Y)]=\iint h(x,y)f_X(x)f_Y(y) ~dx~dy$ if X and Y are independent.*

erm... good luck

How did you get your equality (a) ?

3. Originally Posted by kuywam
Hello
I am Ph.D. student studying electrical engineering.
I have a question related with cdf of function of two random variables.
The question is attached in this post as pdf file.

As an addition to Moo's post, I think what you were looking for is the following formula: (looks very much like (b), but this one is correct)

$P\left(\frac{X}{1+Y}\leq z\right)=$ $P(X\leq z(1+Y))=\int_0^{\infty} P(X\leq z(1+y))f_Y(y)dy = \int_0^\infty F_X(z(1+y))f_Y(y)dy$

(assuming X and Y are independent)

4. ## Thank you so much...^^

Thank you, moo...
Thank you, Laurent...
I will not forget your kind help...^^

5. Originally Posted by kuywam
Thank you, moo...
Thank you, Laurent...
I will not forget your kind help...^^
An acknowledgement (or footnote) in your thesis would be sufficient.