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I've spent so long, and I think I'm missing something super obvious.

The question: Let $X_1,\ldots,X_5$ be five independent variables from the exponential distribution with mean $2$. Write the pdf of $T= X_1+\cdots+X_5$

First off I realize the $n=5$ seems really small, but I haven't learned anything other than the CLT to deal with non-Normal samples. Assuming this works, I can calculate $\lambda$ from mean $=1/\lambda$, and use that to calculate variance $=1/\lambda^2$. Applying the theorem, I then have the distribution as $\approx N(2,4/5)$.

At this point I don't see how to continue. I know the pdf for Normal distributions, but I don't understand the meaning of T and why it is the sum of the random variables. It's got to be obvious, but I'm not seeing it.

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  • $\begingroup$ It's $\displaystyle \frac 1 {\Gamma(5)} \left(\frac x \lambda\right)^{5-1} e^{-x/\lambda}\,\left(\frac{dx}\lambda\right)$ for $x>0$, i.e. a certain Gamma distribution. If I had more time write now, I'd post an answer explaining this. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Jun 12, 2015 at 21:39
  • $\begingroup$ Thanks. I completely missed the fact that Gamma distributions model this scenario, and not CLT. I see that I should use $Gamma(n,\lambda)$ which is what you showed. $\endgroup$
    – Tim
    Jun 12, 2015 at 22:24

1 Answer 1

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It can be proven either by induction using the convolution formula or by use of the moment generating functions that the sum of independent gamma random variables with the same scale/rate parameter is itself gamma distributed with shape equal to the sum of the shape parameters.

To put that in math, pick your favourite definition of your gamma distribution, be it with the rate or scale parameterization. Then assume you have a random sample $X_1, X_2, \ldots, X_n$ from a $Gamma\left( \alpha,\beta \right)$ distribution and define $T=\sum_{i=1}^n X_i$. Then by the above statement $T\sim Gamma \left( n\alpha, \beta \right)$. Notice that this follows because of the same shape parameters but this need not be the case. As long as the summands are independent Gamma random variables with same scale/rate parameter their sum will also be a Gamma random variable.

The exponential distribution is a special case of the broad Gamma family of course and this may be seen by writing out the pdf of the Gamma distribution. Here is the scale parameterization

$$f_X (x) =\begin{cases} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{\alpha-1} e^{-\frac{x}{\beta}} & 0<x<\infty \\ 0 & \text{otherwise} \end{cases}$$

If you now put $\alpha=1$, you will get the exponential pdf, so the properties of the Gamma distribution carry to this case as well.

As per your CLT approach, I would say that your sample is too small to allow for a reasonable approximation. The exponential is a skewed distribution and the skewness does not go away if you sum just five random variables. In fact, this is how the distribution of your sum looks like for the case of a "standard" exponential distribution with $\beta=1$.

enter image description here

The actual shape will also depend on the $\beta$ which controls the amount of skeweness but you get the idea.

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  • $\begingroup$ Thanks for the detailed answer. It definitely felt wrong using CLT, I just didn't see it as a Gamma distribution. $\endgroup$
    – Tim
    Jun 12, 2015 at 23:01

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