My Math Forum Marginal Distribution of Multiple Random Variables

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November 24th, 2010, 05:00 PM   #1
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Marginal Distribution of Multiple Random Variables

Hi all,

Does anybody know how to work with marginal pdfs? If you do, please take a look at the attached problem. Also, if you can, please guide me in the right direction. I have a slight idea of what to do with the problem but not entirely sure.

I know for part a, you obtain the marginal distribution of U and marginal distribution of V (pdf of U and pdf of V) by a single integration of the function. For example, the pdf of U, fU(u), is found by integrating the function f(u,v)du by its respective limits. The same case applies for the pdf of V.

Am I correct so far?

As for part b, finding the CDF of (U, V), do I do a double integration, i.e. FU,V(u,v) = f(u,v)dudv? I am unclear on this part as well as c and d. So if you can offer some insight, that would be great. Thanks for looking.

knp
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 November 24th, 2010, 05:06 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Marginal Distribution of Multiple Random Variables Yes and yes. If they are independent, then the pdf will be the product of the marginal pdfs.
 November 24th, 2010, 10:56 PM #3 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Marginal Distribution of Multiple Random Variables Jason er... DLowry is right. $f_{\small{U}}(u)=\int_{-\infty}^{+\infty}f(u,v)dv=e^{-u}$, $f_{\small{V}}(v)=\int_{-\infty}^{+\infty}f(u,v)du=e^{-v}$, $f(u,v)=f_{\small{U}}(u)f_{\small{V}}(v)$ [color=#FF0000]for all U and V[/color]. So, they are independent. For part b). Let the CDF be $D(u,v)=P\{U\leq u, V\leq v\}$ , we get $D(u,v)=P\{U\leq u, V\leq v\}=\int_{-\infty}^{v}\int_{-\infty}^{u}f(x,y)dxdy=\int_{0}^{v}e^{-y}dy\int_{0}^{u}e^{-x}dx=(1-e^{-u})(1-e^{-v})=1+e^{-u-v}-e^{-u}-e^{-v}$. For part c). $P\{U=>2\}=1-P\{U\leq2\}=1-\int_0^{2} f_{\small{U}}(u)du=1-\int_0^{2} e^{-u}du=e^{-2}$ . $P\{U+V=>2\}=1-P\{U+V\leq2\}=1-\iint_{\small{u+v\leq2}\\u\geq0\\v\geq0}f(u,v)dudv =1-\int_0^2dv\int_{0}^{2-v}f(u,v)du=1-\int_0^2dv\int_{0}^{2-v}e^{-u-v}du=1-\int_0^2dv\int_{0}^{2-v}e^{-u-v}du=1-\int_0^2(e^{-v}-e^{-2})dv=3e^{-2}$ $P\{U\leq v_0\}=\int_0^{v_0} e^{-u}du=1-e^{-v_0}$

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