# Expected Value of Inverse Gamma Distribution

• Yagoda
In summary: Thanks for the help!In summary, the student is having trouble solving a problem involving the gamma function. They know how to approach the problem, but are having trouble getting the answer correct. They have replaced the variable t with 1/y in order to find the expected value, but ran into a problem with having the b^{a-2} pulled outside the integral.
Yagoda

## Homework Statement

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

## Homework Equations

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

## 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.
\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*}

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?

Yagoda said:

## Homework Statement

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

## Homework Equations

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

## 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.
\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*}

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.

Do you have a suggestion for what to use for the substitution? I also tried substituting in $t = \frac{1}{by}$ so that I would just have $e^{-t}$ in the integrand, but I again ran into the problem of having $b^{a-2}$ pulled outside the integral rather than the $b^{a-1}$ that I am supposed to have.

Oh, never mind I think I see where I made my mistake. Forgot about that extra b when substituting dt.

## 1. What is the inverse gamma distribution?

The inverse gamma distribution is a probability distribution that is used to model the probability of a random variable taking on a specific value. It is the inverse of the gamma distribution, which is commonly used in statistics to model positive, skewed data.

## 2. How is the expected value of the inverse gamma distribution calculated?

The expected value of the inverse gamma distribution is calculated by taking the inverse of the shape parameter (alpha) divided by the scale parameter (beta). This can also be expressed as beta divided by alpha-1.

## 3. What does the expected value represent in the inverse gamma distribution?

The expected value represents the mean or average value of the random variable being modeled by the inverse gamma distribution. It is the value that is most likely to occur.

## 4. How is the expected value related to the shape and scale parameters of the inverse gamma distribution?

The expected value is directly affected by the shape and scale parameters of the inverse gamma distribution. As the shape parameter increases, the expected value decreases, while an increase in the scale parameter leads to an increase in the expected value.

## 5. What are some applications of the inverse gamma distribution in science?

The inverse gamma distribution is commonly used in science to model data that is positively skewed, such as in physics, engineering, and finance. It is also used in Bayesian statistics to model the uncertainty of a parameter. Additionally, it has applications in the field of reliability, where it is used to model the time until failure of a product or system.

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