text: # Super Tough Probability Question
  
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  • Nov 25th 2009, 08:25 PM
  Porter1
  Super Tough Probability Question
  ........................
  • 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.