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- May 3rd 2009, 10:22 PMJason2009Independent exponential R.V.
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- May 3rd 2009, 11:49 PMCaptainBlack
The mean is:

$\displaystyle E(Y)=E(a(b+c))=\frac{1}{8}\int_0^{\infty}\int_0^{\ infty}\int_0^{\infty} a(b+c) e^{-a/2}e^{-b/2}e^{-c/2} ~da~db~dc $

which you should be able to do easily (using $\displaystyle \overline{a}=\overline{b}=\overline{c}=2$ )

Then You will need:

$\displaystyle E(Y^2)=E([a(b+c)]^2)=\frac{1}{8}\int_0^{\infty}\int_0^{\infty}\int_ 0^{\infty} [a(b+c)]^2 e^{-a/2}e^{-b/2}e^{-c/2} ~da~db~dc $

(use your knowlege of $\displaystyle \text{var}(a)$ , $\displaystyle \text{var}(b)$ and $\displaystyle \text{var}(c)$ as an aide to evaluating this, you should not need to do any actual integrals that way)

$\displaystyle \text{var}(Y)=E(Y^2)-[E(Y)]^2$

CB - May 4th 2009, 03:40 AMJason2009
Thanks ,CaptainBlack.

What kind of the property did you use? - May 4th 2009, 05:14 AMCaptainBlack
If $\displaystyle X$,$\displaystyle Y$ and $\displaystyle Z$ are independedent RVs then their joint density:

$\displaystyle

p(x,y,z)=p(z)p(y)p(z)

$

where $\displaystyle p(x)$, $\displaystyle p(y)$ and $\displaystyle p(z)$ are the marginal densities of $\displaystyle X$, $\displaystyle Y$ and $\displaystyle Z$ respectivly.

CB