What is the expectation of $X\sim \text{Exp}(\theta,1)$?

Question

Following this question:Definition of $\text{Exp}(\theta,1)$ distribution.

The pdf of $X$ is given as $f(x,\theta)=e^{-(x-\theta)}$, $x>\theta$.

My question is, what is the expectation of $X$?

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My work is that $$ E[X]=e^\theta\int_\theta^\infty xe^{-x}dx=-e^\theta[\theta e^{-\theta}- e^{-\theta}]=1-\theta. $$

Answer 1

It is much easier to recognize that $X$ is a location-transformed exponential distribution with location parameter $\theta$. In other words, $Y = X - \theta$ is exponentially distributed with mean $1$. How do we know this? Well, because if $X$ has density $f_X(x) = e^{-(x-\theta)}$, then $Y = X - \theta$ will have density $$f_Y(y) = f_X(y+\theta) = e^{-(y+\theta-\theta)} = e^{-y}.$$ So we have $$\operatorname{E}[X] = \operatorname{E}[Y + \theta] = \operatorname{E}[Y] + \theta = 1 + \theta.$$ In fact, this reasoning easily generalizes to any location-scale transformation of a random variable: if $X$ has mean $\mu$ and variance $\sigma^2$, then $Y = aX + b$ will have mean $a \mu + b$ and variance $a^2 \sigma^2$.

If we must do the integral calculation, $$\begin{align} \operatorname{E}[X] &= \int_{x=\theta}^\infty x e^{-(x-\theta)} \, dx \\ &= \int_{y=0}^\infty (y+\theta) e^{-y} \, dy & (\text{substitute } x = y + \theta, \; dx = dy) \\ &= \bigl[ (y+\theta)(-e^{-y}) \bigr]_{y=0}^\infty - \int_{y=0}^\infty (-e^{-y}) \, dy & (\text{IBP; } u = y + \theta, \; dv = e^{-y} \, dy) \\ &= \theta - \bigl[e^{-y}\bigr]_{y=0}^\infty \\ &= \theta + 1. \end{align}$$

Answer 2

No, this is not correct. You have that for $X\sim Exp(\theta,\lambda)$,

$$ E[X] = \int_{\theta}^{\infty}xf_X(x)dx = \lambda e^{\lambda\theta}\int_{\theta}^{\infty}xe^{-\lambda x}dx $$

Now, you solve the integral by partial integration (where you went wrong), which written out gives

$$ \int_{\theta}^{\infty}xe^{-\lambda x}dx = \left[x\left(-\frac{e^{-\lambda x}}{\lambda} \right) \right]_\theta^{\infty} + \frac{1}{\lambda}\int_{\theta}^{\infty} e^{-\lambda x}dx = \frac{\theta e^{-\lambda \theta}}{\lambda}+\frac{e^{-\lambda\theta}}{\lambda^2} = \frac{e^{-\lambda\theta}}{\lambda^2}\left(\lambda\theta+1 \right) $$

Plugging this into the expectation value gives

$$ \lambda e^{\lambda\theta}\int_{\theta}^{\infty}xe^{-\lambda x}dx = \lambda e^{\lambda\theta}\frac{e^{-\lambda\theta}}{\lambda^2}\left(\lambda\theta+1 \right) = \frac{\lambda\theta+1}{\lambda} $$

So for $\lambda=1$, you get $E[X] = \theta+1$.