If X is a random variable with the density function $f(x)=\frac{e^{-|x|}}{2}$, what is the CDF of X?
My first inclination is to take $\int_0^\infty \mathrm{e}^{x}/2\,\mathrm{d}x$ and $\int_{-\infty}^0 \mathrm{e}^{-x}/2\,\mathrm{d}x$.
Is this all?
[Math] CDF of a density function with absolute value.
probabilityprobability distributions
Related Solutions
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 $$\frac{1}{2b}e^{-\frac{|x-\theta|}{b}}.$$
Expected Value
In general, the expected value is determined by this following expression
$$\mathrm{E}(X) = \int_{-\infty}^{\infty} xf(x)\,dx$$
where $f(x)$ is the probability density function. For your problem, the expected value is
$$\begin{aligned} \mathrm{E}(X) &= \int_0^1 x\cdot 3x^2\,dx\\ &= \int_0^1 3x^3\,dx\\ &= \left.\dfrac{3}{4}x^4\right\vert_{x = 0}^{x = 1}\\ &= \dfrac{3}{4} \end{aligned}$$
Variance
Recall that the variance is
$$\mathrm{Var}(X) = \mathrm{E}(X^2) - (\mathrm{E}(X))^2$$
We already know $\mathrm{E}(X)$. We then need to compute $\mathrm{E}(X^2)$, which is
$$\begin{aligned} \mathrm{E}(X^2) &= \int_0^1 x^2\cdot 3x^2\,dx\\ &= \int_0^1 3x^4\,dx\\ &= \left.\dfrac{3}{5}x^5\right\vert_{x = 0}^1\\ &= \dfrac{3}{5} \end{aligned}$$
So
$$\mathrm{Var}(X) = \dfrac{3}{5} - \left(\dfrac{3}{4} \right)^2 = \dfrac{3}{80}$$
Standard Deviation
Thus, the standard deviation is
$$\sigma = \sqrt{\mathrm{Var}(X)} = \sqrt{\dfrac{3}{80}}$$
Best Answer
Assuming that $$f(x)=\frac{e^{-|x|}}{2}$$ which is the Laplace distribution, your idea of splitting the function is correct! But, the cdf is itself a function not a single number! That means that your integration limits should depend on x. Therefore, you have to disriminate cases according to the value of x: