Expected value

**dwsmith** · April 9th 2011, 10:45 AM

$\text{E}[L]=\text{E}[30X+2X^2]=30\text{E}[X]+2\text{E}[X^2]$

This is a Gamma Distrib with alpha = 3.5 and beta = 1.5

$\displaystyle p(x)=\frac{1}{\Gamma(3.5)1.5^{3.5}}x^{2.5}e^{-x/1.5^{-1}}$

$\displaystyle\lim_{b\to\infty}\left[\frac{30}{\Gamma(3.5)1.5^{3.5}}\int_0^bx^{3.5}e^{-x/1.5^{-1}} \ dx+\frac{2}{\Gamma(3.5)1.5^{3.5}}\int_0^bx^{4.5}e^ {-x/1.5^{-1}} \ dx\right]$

$\displaystyle =\frac{30\Gamma(4.5)}{\Gamma(3.5)1.5^{8}}+\frac{2\ Gamma(5.5)}{\Gamma(3.5)1.5^{9}}$

$\displaystyle =\frac{30\cdot 3.5}{1.5^8}+\frac{2\cdot 4.5\cdot 3.5}{1.5^9}=4.91632$

The book says the solution is 228.4 though. I don't understand why I am wrong.

**TKHunny** · April 9th 2011, 12:26 PM

You seem a little confused on some notation and some algebra.

1) $x/1.5^{-1}$ ? Too much denominatoring going on in there. Try $x/1.5$ .

2) Just a couple of misconstrued items. You should get:

$30\cdot \frac{\Gamma(4.5)\cdot 1.5}{\Gamma(3.5)} + 2\cdot \frac{\Gamma(5.5)\cdot 1.5^{2}}{\Gamma(3.5)} = 30\cdot 3.5 \cdot 1.5 + 2\cdot 4.5 \cdot 3.5 \cdot 1.5^{2} = 228.375$

This is a tough place to be having an algebra breakdown. Be more careful.

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