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.
• 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