# How is this implied by the properties of the exponential, gamma, and $\chi^2$ distributions?

Let's say we have the random variables $$X_1, \dots, X_p$$. Furthermore, say that these random variables are a random sample from a PDF of the form

$$f_\tau (x) = \begin{cases} \tau x^{\tau-1}, & 0 < x < 1, \tau > 0 \\ 0, &\text{otherwise} \end{cases}$$

Now define $$Y_i = - \log(X_i)$$. If I'm not mistaken, this means that $$Y_i$$ has an exponential distribution.

I am then told that

$$E \left[ \dfrac{1}{2 \tau \Sigma_{i = 1}^p Y_i} \right] = \dfrac{1}{2(p - 1)}$$

and

$$\text{Var} \left( \dfrac{1}{2 \tau \Sigma_{i = 1}^p Y_i} \right) = \dfrac{1}{4(p - 1)^2(p - 2)}$$

Apparently, this is implied by the properties of the exponential, gamma, and $$\chi^2$$ distributions. However, I have no idea how this is the case. Trying to 'fill in the blanks' here, the denominators $$2(p - 1)$$ and $$4(p - 1)^2(p - 2)$$ look like something you'd get related to the DOF for a $$\chi^2$$ random variable, but, as I said, so much information is left out here that I can't follow the reasoning. How are these results implied by the properties of the exponential, gamma, and $$\chi^2$$ distributions? What exactly is the reasoning used here?

• You are provided complete information about this situation--nothing is left out. There are many ways to derive these results. One useful starting point might be to determine the expectation and variance of the Inverse Gamma distribution. You then only need to know that sums of iid Exponential variables have Gamma distributions.
– whuber
Apr 9 '21 at 13:47
• @whuber Hmm, how does the inverse gamma distribution relate to this? Apr 9 '21 at 13:50
• Given that the $\chi^2$ distribution is a special case of Gamma distribution, you can focus on Gammas only. Apr 9 '21 at 13:53
• @ThePointer Do you know the distribution of $\sum_i Y_i$ or its inverse? Apr 9 '21 at 13:53
• Apr 9 '21 at 14:43