Expectation of a square root of a sample mean

Let $X_i$ be iid exponential random variables. I want to calculate $Var\left(\sqrt{\bar{X}/6}\right).$

The idea I had to simplify this is expressing it as $$Var\left(\sqrt{\frac{\bar{X}}{6}}\right) = E\left(\frac{\bar{X}}{6}\right) - \left[E\left(\sqrt{\frac{\bar{X}}{6}}\right)\right]^2.$$

This still leaves me with the problem of solving $E(\sqrt{\bar{X}/6})$. How do I deal with the square root inside the expectation?

• It is not going to be a real number because a normal random variable takes on negative values. – user3903581 Apr 2 '17 at 1:08
• @user3903581 Thank you, let me amend the question to make it more practical – kingledion Apr 2 '17 at 1:09
• Is the sample size $n$ or 6? Anyway, you may be able to use the fact that the sum of Exponentials is a Gamma distribution ... – wolfies Apr 2 '17 at 6:30

The sample mean from an exponential($\lambda$) sample is gamma($n$, $\lambda/n$) distributed.

It's a fact that the square root of $\overline{X}$ therefore has a generalized gamma distribution with parameters $p=2$, $d=2n$, and $a = \sqrt{\lambda/n}$, so it's mean is

$$a \frac{ \Gamma( (d+1)/p )}{\Gamma(d/p)} = \sqrt{\lambda/n} \cdot \frac{ \Gamma( (2n+1)/2)}{ \Gamma(n)}$$

Divide this by $\sqrt{6}$ and there's your answer.

• Is $\lambda$ standing for a scale parameter in your formulation? – Alecos Papadopoulos Apr 2 '17 at 14:03
• Yes. If $\lambda$ is supposed to be the rate, replace $\lambda$ with $1/\lambda$ in the final answer. – gammer Apr 2 '17 at 14:44

Say the parameter to each distribution is $\lambda$. First, $$E\left[\frac{\bar{X}}{6}\right] = [6\lambda]^{-1}$$ by linearity. Then, because $Y = \sum_i X_i \sim \text{Gamma}(n,1/\lambda)$ \begin{align*} E\left(\sqrt{\frac{\bar{X}}{6}}\right)&= \int \sqrt{\frac{y}{6n}} \frac{\lambda^n }{\Gamma(n)}y^{n-1}e^{-y\lambda} dy \\ &= (6n)^{-1/2}\frac{\lambda^n}{(n-1)!} \int y^{n + 1/2 - 1}e^{-y\lambda}dy \\ &= (6n)^{-1/2}\frac{\lambda^n}{(n-1)!} \left[\frac{\Gamma(n+1/2) \lambda^{-(n+1/2)} }{\Gamma(n+1/2)\lambda^{-(n+1/2)}}\right] \int y^{n + 1/2 }e^{-y\lambda}dy \\ &= (6n)^{-1/2}\frac{\lambda^n}{(n-1)!} \Gamma(n+1/2) \lambda^{-(n+1/2-1)} \\ &= (6n)^{-1/2} \lambda^{1/2} \frac{\Gamma(n+1/2) }{\Gamma(n) } \end{align*}

The Exponential distribution is a special case of a Gamma ($G$) distribution. If $X \sim \text{Exp}(\lambda)$ where $\lambda$ is the rate parameter (so ($E(X) = 1/\lambda)$, then (shape -scale parametrization)

$$\sum_{i=1}^n X_i \sim G(n, 1/\lambda) \implies \bar X = \frac 1n \sum_{i=1}^n X_i \sim G(n, 1/n\lambda)$$

Now, $$Z = \sqrt {\bar X} \implies \bar X = Z^2 \implies \frac {\partial \bar X}{\partial Z} = 2Z$$

Then by the change-of-variable formula we have that the density of $Z$ is

$$f_Z(z) = \left |\frac {\partial \bar X}{\partial Z}\right| \cdot f_{\bar X}(z)$$

$$\implies f_Z(z) = \frac {2z}{\Gamma(n)\cdot (1/n\lambda)^n}\cdot (z^2)^{n-1} \cdot\exp \{- z^2/(1/n\lambda)\}$$

$$\implies f_Z(z) = \frac {2}{\Gamma(n)\cdot (1/n\lambda)^n}\cdot z^{2n-1} \cdot\exp \{- z^2/(1/n\lambda)\}$$

Set $p\equiv 2,\;\; n \equiv d/p \equiv d/2,\;\; a\equiv (1/n\lambda)^{1/2}$. Then we can write

$$f_Z(z) = \frac {p}{\Gamma(d/p)\cdot a^d}\cdot z^{d-1} \cdot\exp \{- z^2/a^2\}$$

This is the density of the Generalized Gamma distribution, and so we have

$$E(Z) = E\left[\sqrt{\bar X}\right] = a\frac {\Gamma[(d+1)/p]}{\Gamma(d/p)}$$

and reverting back to the original parameters we obtain

$$E\left[\sqrt{\bar X}\right] = \frac{1}{\sqrt{n\lambda}}\cdot \frac {\Gamma[n+(1/2)]}{\Gamma(n)}$$