Finding the Expectation and Variance, given the distribution function and density function for a continuous random variable

Question

For this density function of a continuous random variable, X:

$$f(x) = \begin{cases} c & \mbox{for } -1 ≤ x ≤ 1\\ \tfrac{c}{x^4} & \mbox{for } |x| > 1\end{cases}$$

and calculating its distribution function:

$$F(x) = \begin{cases} \int_{1}^{\infty} \tfrac{3}{8x^4} dx & \text{if } x>1 \\ \int_{-1}^{1} \tfrac{3}{8} dx & \text{if } -1 ≤ x ≤ 1 \\ \int_{-\infty}^{-1} \tfrac{3}{8x^4} & \text{if } x < -1\end{cases}$$

I am now unsure how to calculate the Expectation, Variance and Median.

Every time I try to using $\int_{}^{} xf(x) dx$ for each of the intervals, I seem to get $EX = 0$

Any help will be greatly appreciated. Thanks

Answer 1

Lets be more precise about things here.

Firstly, the integrals you have written down for the distribution function $F(x)$ (which I assume you mean the Cumulative Distribution Function) are simply numbers. For example, $$\int_1^\infty \frac{3}{8x^4} dx = \frac 18$$

is not something that depends on $x$. Instead, recall that the function $F(x)$ is defined as

$$F(x) = \int_{-\infty}^x f(s)ds \tag{1}$$

whereas the function $f(x)$ satisfies

$$\int_{-\infty}^\infty f(s)ds = 1 \tag{2}$$

I assume you have already used $(2)$ and found the constant $c = \frac 38$

Thus when $x<-1$, using $(1)$, we have

$$F(x) = \int_{-\infty}^x \frac{3}{8s^4}ds = -\frac{1}{8x^3}$$

I will leave you to do the other two cases, but you should find that

$$F(x) = \begin{cases} -\frac{1}{8x^3} & x \leq -1 \\ \frac 38 x + \frac 12 & -1<x<1 \\ 1-\frac{1}{8x^3} & x>1 \end{cases}$$

Next, as you have noted, we calculate the expectation using the formula

$$\Bbb E (X) = \int_{-\infty}^\infty xf(x)dx$$

But of course, for our function $f(x)$ we will have to split the domains of integration:

\begin{align} \Bbb E (X) & = \int_{-\infty}^\infty xf(x)dx \\ & = \int_{-\infty}^{-1} x \cdot \frac{3}{8x^4} dx + \int_{-1}^1 x \cdot \frac 38 dx + \int_1^\infty x \cdot \frac{3}{8x^4} dx \\ & = \int_{-\infty}^{-1} \frac{3}{8x^3} dx + \int_{-1}^1 \frac {3x}{8} dx + \int_1^\infty \frac{3}{8x^3} dx \\ & = -\frac{3}{16} + 0 + \frac{3}{16} \\ & = 0 \end{align}

Indeed we find that the expected value is $0$, which is not surprising due to the fact that $f(x)$ is an even function.

Next, the formula for the variance is

$$\text{Var}(X) = \Bbb E(X^2) - \big[\Bbb E(X)\big]^2$$

We already know that $\Bbb E(X)=0$, so all that is left is to calculate $\Bbb E(X^2)$. This is done in a similar fashion as before:

\begin{align} \Bbb E (X^2) & = \int_{-\infty}^\infty x^2f(x)dx \\ & = \int_{-\infty}^{-1} x^2 \cdot \frac{3}{8x^4} dx + \int_{-1}^1 x^2 \cdot \frac 38 dx + \int_1^\infty x^2 \cdot \frac{3}{8x^4} dx \\ & = \int_{-\infty}^{-1} \frac{3}{8x^2} dx + \int_{-1}^1 \frac {3x^2}{8} dx + \int_1^\infty \frac{3}{8x^2} dx \\ & = \frac{3}{8} + \frac 14 + \frac{3}{8} \\ & = 1 \end{align}

It follows that the variance is

$$\text{Var}(X) = \Bbb E(X^2) - \big[\Bbb E(X)\big]^2 = 1 - (0)^2 = 1$$

Finally, the Median amounts to finding the value of $x$ such that $F(x)=\frac 12$. This is easily found to be $x = 0$.