# Finding the Mean and Variance of this distribution

Given the following:

$$Y|\Lambda=\mathrm{Poisson(\Lambda)}$$ and $$\Lambda=\mathrm{Gamma}(\alpha, \beta)$$

Find the distribution, mean and variance of $$Y$$.

After a lot of work, I was able to find the distribution of $$Y$$ which is:

$$f_Y(y)={{\Gamma(y+\alpha)e^{(-y/\beta)}}\over{y!\Gamma(\alpha)\beta^\alpha}}$$

Now using this distribution, I wanted to find the mean and variance. I tried integrating the above results but I couldn't find a closed form for the product $$\Pi_{i=1}^n{(y+\alpha-i)}$$ and from their the integration breaks down.

I would appreciate if someone could provide me a way to calculate the mean and varaiance of Y. Thanks

• $Y$ should have a negative-binomial distribution (see en.wikipedia.org/wiki/…), from which the moments follow. Jan 19, 2020 at 16:21
• Yes the question tells us to show that it is a negative binomial distribution but could you please show me how you came to this conclusion through the distribution of Y? As my distribution doesn't consist of probability but the exp function Jan 19, 2020 at 16:34
• Did you see the derivation on the wiki page? You made a mistake somewhere. Jan 19, 2020 at 16:38
• Ah yes. I see my mistake. Thank you very much Jan 19, 2020 at 17:27
• You can get displayed equations by enclosing them in double instead of single dollar signs. This is particularly helpful when you're mixing fractions and exponents, which tend to become hard to read the way you typeset them. Jan 19, 2020 at 17:44

I solved this question by setting $$\beta = (p/(1-p))$$. I then followed the derivation of the Negative Binomial distribution as shown in this link. Since I wanted an answer in terms of $$\beta$$, I simply solved the above expression for $$p$$ and replaced $$\beta$$ with that expression, giving me Y~NB($$\alpha$$, $$\beta/(1+\beta)$$). From there the mean and variance can easily be found.