# Variance of $\frac{X_i}{\theta^2} -\frac{1}{\theta}$ in an exponential distribution

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 $$f(x;\theta) = \frac{1}{\theta} \exp^{-\frac{x}{\theta}}$$.

Thanks.

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.

• 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 – Ricky_Nelson Apr 29 at 2:41
• 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. – javierazcoiti Apr 29 at 3:33
• Ah, got it, thanks! – Ricky_Nelson Apr 29 at 3:44
• You're welcome. And sorry for saying "he is using" and "his case". I mean, you :) – javierazcoiti Apr 29 at 4:20