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While I tagged this question homework, it's not really homework. I'm really studying for a test and don't understand how to do something. I have the answer to the question, so the important part is for my to understand how to get there.

X has distribution Gamma(2,4) X = Y-3 Find: P(Y>11|Y>7) Which I know is equals to:

(1-P(X<8))/(1-P(X<4)) = (1-Fx(8))/(1-Fx(4))

I have been told that Fx(a) = 1-e^(-a/4)-(a/4)*e^(-a/4)

So I can clearly solve the problem now it's just algebra. But how do we know what Fx(a) equals???


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up vote 3 down vote accepted

$F_x(a)$ is the cumulative distribution function of a gamma$(2,4)$ distribution. This is given by $$F_x(a) = \int_0^a \frac{t e^{-t/4}}{(4^2)\Gamma(2)} dt.$$ Integrating (by parts) yields the given expression for $F_x(a)$, since $\Gamma(2) = (2-1)! = 1$.

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O.K. so it seems as though the best way in general to do this is know that P(x<a) = integral of the probability density function evaluated at a? – kralco626 Dec 15 '10 at 23:24
@kralco626: In general, yes, although when you find the integral make sure you include the lower bound as well as the upper bound. – Mike Spivey Dec 15 '10 at 23:34
how do i know what the bounds are? is this because gamma is only for x>0. Also is Γ(x) always (x-1)!? – kralco626 Dec 15 '10 at 23:58
@kralco626: Yes to your second question, which implies the answer to the first. As far as the third, when $x$ is a non-negative integer, $\Gamma(x) = (x-1)!$. Otherwise, see the definition of the gamma function: – Mike Spivey Dec 16 '10 at 0:04
thanks so much! – kralco626 Dec 16 '10 at 0:13

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