Density function with absolute value Let $X$ be a random variable distributed with the following density function:
$$f(x)=\frac{1}{2} \exp(-|x-\theta|) \>.$$
Calculate: $$F(t)=\mathbb P[X\leq t], \mathbb E[X] , \mathrm{Var}[X]$$
I have problems calculating $F(t)$ because of the absolute value. I'm doing it by case statements but it just doesn't get me to the right answer.
So it gets to this:
$$
\int_{-\infty}^\infty\frac{1}{2} \exp(-|x-\theta|)\,\mathrm dx $$
 A: The first approach we take, though correct, is not the best one, and we later describe a better approach.  
Suppose first that $x \ge \theta$.  By symmetry, the probability that $X\le \theta$ is $\frac{1}{2}$.  So if $x\ge \theta$, then 
$$P(X\le x)= \frac{1}{2}+\int_{\theta}^x \frac{1}{2}e^{-(t-\theta)}\,dt.$$
The integration can be done by pulling out the $e^\theta$, but I prefer to make the substitution $u=t-\theta$.  The integral becomes
$$\int_{u=0}^{x+\theta} \frac{1}{2}e^{-u}du,$$
which evaluates to
$$\frac{1}{2}(1-e^{-(x-\theta}).$$
Adding the $\frac{1}{2}$ for the probability that $X\le \theta$, we find that for $x\ge\theta$, $F_X(x)=1-\frac{1}{2}e^{-(x-\theta)}.$
For $x<\theta$, we need to find
$$\int_{-\infty}^x \frac{1}{2}e^{t-\theta}dt.$$
The integration is straightforward. We get that $F_X(x)=\frac{1}{2}e^{x-\theta}$ whenever $x <\theta$.  We could go on the find the mean and variance by similar calculations, but will now change approach.
Another approach: The $\theta$ is a nuisance. Let's get rid of it. So let $Y=X-\theta$.  Then $P(Y\le y)=P(X\le y-\theta)$.  This is 
$$\int_{-\infty}^{y-\theta} \frac{1}{2}e^{-|t-\theta|}dt.$$
Make the change of variable $w=t-\theta$. We find that our integral is
$$\int_{w=-\infty}^y \frac{1}{2}e^{-|w|}dw.$$
What this shows is the intuitively obvious fact that $Y$ has a distribution of the same family  as the one for $X$, except that now the parameter is $0$.  We could now repeat our integration work, with less risk of error.  But that would be a waste of space, so instead we go on to find the expectation of $X$.
Since $X=Y+\theta$, we have $E(X)=E(Y)+\theta$. On the assumption that this expectation exists, by symmetry $E(Y)=0$, and therefore $E(X)=\theta$. 
Next we deal with $\text{Var}(X)$. Since $X=Y+\theta$, the variance of $X$ is the same as the variance of $Y$. So we need to find
$$\int_{-\infty}^\infty \frac{1}{2}w^2e^{-|w|}dw.$$
By symmetry, this is twice the integral from $0$ to $\infty$, so we want
$$\int_0^\infty w^2e^{-w}dw.$$
Integration by parts (twice) handles this problem. To start, let $u=w^2$, and let $dv=e^{-w}dw$.
After a little while, you should find that the variance of $Y$, and hence of $X$, is $2$.
Remark: You can also find the mean and variance of $X$ by working directly with the original density function of $X$, and making an immediate substitution for $x-\theta$.  But defining the new random variable $Y$ is in my view a more "probabilistic" approach. 
Your distribution is a special case of the Laplace distribution, which in addition to a location parameter $\theta$, has a scale parameter $b$. The probability density function is 
$$\frac{1}{2b}e^{-\frac{|x-\theta|}{b}}.$$
A: The very best thing you can do in solving problems such as these is to sketch the given density function first.  It does not have to be a very accurate sketch: if you drew a peak of $\frac{1}{2}$ at $x=\theta$ and decaying curves on either side, that's good enough! 
Finding $F_X(t)$:


*

*Pick a number $t$ that is smaller than $\theta$ (where that peak is) and remember that $F_X(t)$ is just the area under the exponential curve to the left of $t$. You can find this area by integration.  

*Think why it must be that $F_X(\theta) = \frac{1}{2}$.  

*Pick a $t > \theta$.  Once again, you have to find $F_X(t)$ which is 
the area under the density to the left of $t$.  This is clearly the area to the left of $\theta$ (said area is $\frac{1}{2}$, of course!) plus the area under the curve between $\theta$ and $t$ which you can find by
integration.  Or you can be clever about it and say that the area to the right of
$t = \theta + 5$ must, by symmetry, equal the area to the left of $\theta - 5$, which you found previously. Since the total area is $1$, we have $F_X(\theta+5)=1-F_X(\theta-5)$, or  more generally,
$$F_X(\theta + \alpha) = 1 - F_X(\theta - \alpha).$$
Finding $E[X]$:
Since the pdf is symmetric about $\theta$, it should work out that $E[X]=\theta$
but we do need to check that the integral does not work out to be of the undefined form $\infty-\infty$.
A: If $x\ge\theta$ then $|x-\theta|=x-\theta$.
If $x<\theta$ then $|x-\theta| = \theta-x$.
So
$$
\int_{-\infty}^\infty x \frac 1 2 \exp(-|x-\theta|)\,dx = \int_{-\infty}^\theta x\frac 1 2 \exp(\theta-x)\;dx + \frac 1 2 \int_\theta^\infty x \exp(\theta-x)\;dx.
$$
By symmetry, the expected value should be $\theta$ if there is an expected value at all.  And, as it happens, there is.  The only thing that would prevent that is if one of the integrals were $+\infty$ and the other $-\infty$.
If you use the substitution
$$
u = x-\theta, \qquad du = dx,
$$
then what you have above becomes
$$
\frac 1 2 \int_{-\infty}^0 (u+\theta) \exp(u)\;du + \frac 1 2\int_0^\infty (u+\theta)\exp(-u)\;du
$$
This is
$$
\begin{align}
& \frac 1 2 \int_{-\infty}^0 u \exp(u)\;du + \frac 1 2 \int_0^\infty u\exp(-u)\;du + \frac 1 2 \int_{-\infty}^0 \theta \exp(u)\;du + \frac 1 2 \int_0^\infty \theta\exp(-u)\;du \\  \\
& = \frac 1 2 \int_{-\infty}^\infty u \exp(-|u|)\;du + \theta\int_{-\infty}^\infty \frac 1 2 \exp(-|u|)\;du
\end{align}
$$
The first integral on the last line is $0$ because you're integrating an odd function over the whole line.  The second is $1$ because you're integrating a probability density function over the whole line.
So you're left with $\theta$.
