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Math Help - pdf of function of two random variables (It is little difficult)

  1. #1
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    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.

    Please, give me an answer.
    Attached Files Attached Files
    Last edited by kuywam; November 23rd 2009 at 05:27 AM.
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  2. #2
    Moo
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    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}<br />
P\left(\tfrac{X}{1+Y}\leq z\right)=P(X \leq z(1+Y))<br />
&=P\left(Y\geq \tfrac{X-z}{z}\right) \\<br />
&=\int_{x'=0}^\infty \int_{y'=0}^{\tfrac{x'-z}{z}} f_Y(y')f_X(x') ~dy' ~dx' \quad (*) \\<br />
&=\int_{x'=0}^\infty f_X(x')\left(\int_{y'=0}^{\tfrac{x'-z}{z}}  f_Y(y') ~dy'\right) ~dx' \\<br />
&=\int_{x'=0}^\infty f_Y(y') F_Y\left(\tfrac{x'-z}{z}\right) ~dx \\<br />
&=\mathbb{E}\left[F_Y\left(\tfrac{X-z}{z}\right)\right]<br />
\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) ?
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  3. #3
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    Quote Originally Posted by kuywam View Post
    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.
    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)
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  4. #4
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    Thank you so much...^^

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