# Super Tough Probability Question

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• Nov 25th 2009, 08:25 PM
Porter1
Super Tough Probability Question
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• Nov 26th 2009, 12:42 AM
matheagle
Just do (a) directly with polar co-ordinates.
That's how you prove the normal density integrates to one.
• Nov 26th 2009, 08:00 AM
awkward
Quote:

Originally Posted by Porter1
Hi all, I have this question and I really need your help!

(a) Show that Г(1/2) = √(П) (<-Square root of Pi) by writing
Г(1/2) = ∫(0)->(∞) y^(-1/2) e^(-y) dy
by making the transformation y = (x^2)/2 and using the standard normal density.
[snip]

You can work this out from basics as suggested by MathEagle, but if you are allowed to assume some basic knowledge of the Normal distribution there is a shortcut, which I think is probably what you are intended to do--

After you make the substitution, you should have
$\Gamma(1/2) = \sqrt{2} \; \int_0^\infty \exp(-x^2/2) \, dx$
which is equal to
$\sqrt{2} \cdot \sqrt{2 \pi} \cdot \frac{1}{\sqrt{2 \pi}} \; \int_0^{\infty} \exp(-x^2/2) \, dx$
Based on your knowledge of the Normal distribution, you should know the value of $\frac{1}{\sqrt{2 \pi}} \; \int_0^{\infty} \exp(-x^2/2) \, dx$.

Take it from there.