I am pretty lost on how to even look at this

Let $X$ be identical and independently distributed exponential random variables with parameter $\lambda = 1$ for $i = 1,2,\dots,n$. Then$\sum X_i \ \sim Gamma (n,1)$. Find the probability that the sample mean is within $2$ standard deviations of the true mean, for

  • n=4
  • n=12
  • n=20
  • n=20 using an appropriate approximate distribution

I've thought of this so far, which may be wrong: $$ E(\bar{x}) = 1 \ (which\ I\ figured\ out\ because\ beta=1\ in\ Exp(1)\ )\\ Var(\bar{x}) = \frac{\beta^2}{n} ---> s.d. = \frac{1}{2} $$ And then I would approximate the probabilities with a normal distribution with a $z$-test. I am basically unsure if I am even going through this correctly though, any help would be appreciated.


1 Answer 1


One must understand the difference between the sample mean and the true mean. The former is a statistic, therefore a random variable. The latter is a parameter, thus fixed, and in this case, known.

As you were told, $\sum_{i=1}^n X_i \sim \operatorname{Gamma}(n, 1)$, where each $X_i \sim \operatorname{Exponential}(\lambda = 1)$ is iid. What this statement indicates is that the sample total of $n$ iid observations $(x_1, \ldots, x_n)$ from an exponential distribution with rate $1$ is gamma distributed with shape $n$ and rate $1$. Therefore, consider the sample mean $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ which by virtue of a scale transformation, must also be gamma distributed. Explicitly, and for ease of notation, let $T_n = \sum_{i=1}^n X_i$, so that $\bar X = T_n/n$. Then $$f_{\bar X}(x) = f_{T_n}(nx) \left|\frac{d}{dx}[nx]\right| = nf_{T_n}(nx) = n \frac{(nx)^{n-1} e^{-nx}}{\Gamma(n)};$$ in other words, $$\bar X \sim \operatorname{Gamma}(n, n),$$ where the parametrization is by rate. For a general gamma distribution with shape $n$ and rate $\lambda$, the mean is $n/\lambda$ and the variance is $n/\lambda^2$; so $$\operatorname{E}[\bar X] = 1, \quad \operatorname{Var}[\bar X] = 1/n.$$ You would therefore need to numerically calculate for the first case $n = 4$ $$\Pr[1 - 2\sqrt{1/4} < \bar X < 1 + 2\sqrt{1/4}] = \Pr[0 < \bar X < 2]$$ where $\bar X$ is distributed as above. This can be done, for example, by computing via integration by parts $$\Pr[0 < \bar X < 2] = \int_{x=0}^2 \frac{128}{3} x^3 e^{-4x} \, dx.$$ For $n = 12$, this becomes $$\Pr\left[1 - \tfrac{1}{\sqrt{3}} < X < 1 + \tfrac{1}{\sqrt{3}}\right] = \frac{2^{16} 3^8}{1925} \int_{x=1-1/\sqrt{3}}^{1+1/\sqrt{3}} x^{11} e^{-12x} \, dx.$$ For $n = 20$, a normal approximation is appropriate; we would say that $$Z = \frac{\bar X - \operatorname{E}[\bar X]}{\sqrt{\operatorname{Var}[\bar X]}} = \frac{\bar X - 1}{\sqrt{1/20}} \sim \operatorname{Normal}(0,1)$$ approximately, hence the probability that $\bar X$ is two standard deviations within the mean is simply $$\Pr[|Z| < 2],$$ which we can calculate without any information about $\bar X$--it comes from a standard normal table.

  • $\begingroup$ This makes a lot more sense!!! Thanks for answering like every part too, I did not expect that $\endgroup$
    – ajdawg
    Commented Nov 9, 2018 at 8:46
  • $\begingroup$ Oh but one question regarding Γ(n). Is that just the integral from 0 to infinity of x^(n-1)*e^(-n)dx? or is it something else? $\endgroup$
    – ajdawg
    Commented Nov 9, 2018 at 8:47
  • $\begingroup$ @ajdawg For positive integers $n$, $\Gamma(n) = (n-1)!$, so for example, $\Gamma(4) = 3! = 6$. The integral representation is not necessary because $n$ is a sample size, and the other parts of the calculation do not apply since you are integrating over a finite interval. $\endgroup$
    – heropup
    Commented Nov 9, 2018 at 9:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .