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Expected Value of Inverse Gamma Distribution

  • Thread starter Yagoda
  • Start date
  • #1
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Homework Statement


I have that X is distributed with Gamma(a,b) and that [itex] Y = \frac{1}{X}[/itex]. I found the pdf of Y to be [itex]\frac{1}{\Gamma(a)b^a} \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} [/itex] for y > 0. I need to use this to find the expected value.


Homework Equations


The gamma function is defined as [itex]\Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} \, dt [/itex]


The Attempt at a Solution

I know how to approach the problem, but I am having trouble actually getting the correct answer. Here is what I've done.
[itex] \begin{align*}
E[Y] &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} y \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} \, dy \\
&=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty}\left(\frac{1}{y}\right)^{a} e^{-1/yb} \, dy \\
\text{Let t = 1/y so dt = -1/y^2.} \\
&= \frac{-1}{\Gamma(a)b^a} \int_{\infty}^{0} t^{a-2} e^{-t/b} \, dt \\
&= \frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} \left(\frac{t}{b}\right)^{a-2} b^{a-2} e^{-t/b} \, dt \\
&= \frac{b^{a-2}\Gamma(a-1)}{\Gamma(a)b^a} \\
&= \frac{1}{\Gamma(a-1) b^2}
\end{align*}
[/itex]

But I think I'm not supposed to have that extra b in the denominator. I've gone over the substitution a few times and it seems like raising t to the a-2 power is correct since I've removed two of the (1/y)'s to substitute in for dt. What am I missing?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Helper
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Homework Statement


I have that X is distributed with Gamma(a,b) and that [itex] Y = \frac{1}{X}[/itex]. I found the pdf of Y to be [itex]\frac{1}{\Gamma(a)b^a} \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} [/itex] for y > 0. I need to use this to find the expected value.


Homework Equations


The gamma function is defined as [itex]\Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} \, dt [/itex]


The Attempt at a Solution

I know how to approach the problem, but I am having trouble actually getting the correct answer. Here is what I've done.
[itex] \begin{align*}
E[Y] &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} y \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} \, dy \\
&=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty}\left(\frac{1}{y}\right)^{a} e^{-1/yb} \, dy \\
\text{Let t = 1/y so dt = -1/y^2.} \\
&= \frac{-1}{\Gamma(a)b^a} \int_{\infty}^{0} t^{a-2} e^{-t/b} \, dt \\
&= \frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} \left(\frac{t}{b}\right)^{a-2} b^{a-2} e^{-t/b} \, dt \\
&= \frac{b^{a-2}\Gamma(a-1)}{\Gamma(a)b^a} \\
&= \frac{1}{\Gamma(a-1) b^2}
\end{align*}
[/itex]

But I think I'm not supposed to have that extra b in the denominator. I've gone over the substitution a few times and it seems like raising t to the a-2 power is correct since I've removed two of the (1/y)'s to substitute in for dt. What am I missing?
You have changed variables incorrectly: when you set ##t = 1/y## you will have ##e^{-t^b}## in the integrand, so you cannot 'do' the integral as it stands. You need a different change of variables.
 
  • #3
46
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Do you have a suggestion for what to use for the substitution? I also tried substituting in [itex]t = \frac{1}{by} [/itex] so that I would just have [itex] e^{-t} [/itex] in the integrand, but I again ran into the problem of having [itex]b^{a-2}[/itex] pulled outside the integral rather than the [itex]b^{a-1}[/itex] that I am supposed to have.
 
  • #4
46
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Oh, never mind I think I see where I made my mistake. Forgot about that extra b when substituting dt.
 

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