I read in a book discussing the exponential distribution that the variance of $\frac{X_i}{\theta^2} -\frac{1}{\theta}$ is equal to $\frac{1}{\theta^4}Var(X_i) = \frac{1}{\theta^4}\theta^2$. Can someone please explain how $Var(X_i) = \theta^2$?

Note: It was also given in the book that $X_i$ is a random sample from an exponential distribution with pdf $f(x;\theta) = \frac{1}{\theta} \exp^{-\frac{x}{\theta}}$.



Your exponential distribution has mean $\theta$ and variance $\theta^2$.

If X has variance $\sigma$, then $(aX+b)$ has variance $a^2 σ$. This means, $b$ just translates the distribution but doesn't affect its variance, it just changes its mean.

The same can be applied in your case.

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  • $\begingroup$ Isn't the variance of the exponential distribution $\frac{1}{\theta^2}$ instead of $\theta^2$? For instance, see: statlect.com/probability-distributions/exponential-distribution $\endgroup$ – Ricky_Nelson Apr 29 at 2:41
  • $\begingroup$ It depends how you use the parameter of the exponential in the PDF. He is using 1/θ as the rate, λ. The mean and variance of exponential is 1/λ and 1/λ^2. This is, θ and θ^2 in his case. $\endgroup$ – javierazcoiti Apr 29 at 3:33
  • $\begingroup$ Ah, got it, thanks! $\endgroup$ – Ricky_Nelson Apr 29 at 3:44
  • $\begingroup$ You're welcome. And sorry for saying "he is using" and "his case". I mean, you :) $\endgroup$ – javierazcoiti Apr 29 at 4:20

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