# How does one compute estimate of $\theta$ for this density function?

Currently, I am studying for my exam in Statistics. We use Rice's book, so I am trying to do some of the exercises in this book. One of the exercises is as follows:

Suppose $X_1, \dots, X_n$ are i.i.d. with density function $f(x \mid \theta) = e^{ - (x - \theta) }$, when $x \geq \theta$ , and $f(x \mid \theta ) = 0$ otherwise. Find the method of moments estimate of $\theta$.

I know that, in order to compute the method of moments estimate of $\theta$, one has to equate the theoretical average $\bar{X}$ with the expectation. I don't know how to compute the expectation, though. I was thinking of finding it with a double integral:

$$\mathbb E[f(x\mid\theta)] = \int_{0}^{\infty}\!\int_{\theta}^{\infty} e^{ - (x - \theta ) }\,dx\,d\theta,$$ and then doing something similar in the region left of the y-axis. This, however, did not lead to the correct result.

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An easier way to do this, if you are familiar with common distributions, is to recognize that the density function you were given is almost the density of an Exponential(1) random variable--to be more precise, $X-\theta$ follows an Exponential(1) distribution. So $\mathbb E[ X-\theta] = 1$ (the expectation of an Exponential(1) distribution), and thus $\mathbb E[X] = 1+\theta$, the same result akkkk obtained--but without having to do integration by parts!
$\theta$ is just a parameter.
$$\mathbb{E}_\theta[X] = \int_{-\infty}^\infty xf(x|\theta) dx=\int_{\theta}^{\infty} x e^{ - (x - \theta ) } dx\overset{\text{integration by parts}}{=}1+\theta$$