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.
Please, give me an answer.
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.
Please, give me an answer.
Hello,
(a) is ... weird Oo
First thing : if X and Y are independent, you can write the product $\displaystyle f_X(x')f_Y(y')$
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 :
$\displaystyle \begin{aligned}
P\left(\tfrac{X}{1+Y}\leq z\right)=P(X \leq z(1+Y))
&=P\left(Y\geq \tfrac{X-z}{z}\right) \\
&=\int_{x'=0}^\infty \int_{y'=0}^{\tfrac{x'-z}{z}} f_Y(y')f_X(x') ~dy' ~dx' \quad (*) \\
&=\int_{x'=0}^\infty f_X(x')\left(\int_{y'=0}^{\tfrac{x'-z}{z}} f_Y(y') ~dy'\right) ~dx' \\
&=\int_{x'=0}^\infty f_Y(y') F_Y\left(\tfrac{x'-z}{z}\right) ~dx \\
&=\mathbb{E}\left[F_Y\left(\tfrac{X-z}{z}\right)\right]
\end{aligned}$
For the boundaries in (*), that's using the fact that $\displaystyle P\left(Y\geq \frac{X-z}{z}\right)=\mathbb{E}\left[\bold{1}_{Y\geq \frac{X-z}{z}}\right]$
and that for any measurable function h, $\displaystyle \mathbb{E}[h(X,Y)]=\iint h(x,y)f_X(x)f_Y(y) ~dx~dy$ if X and Y are independent.*
erm... good luck
How did you get your equality (a) ?
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)
$\displaystyle P\left(\frac{X}{1+Y}\leq z\right)=$ $\displaystyle P(X\leq z(1+Y))=\int_0^{\infty} P(X\leq z(1+y))f_Y(y)dy = \int_0^\infty F_X(z(1+y))f_Y(y)dy$
(assuming X and Y are independent)