Ytukyg

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

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

and calculating its distribution function:

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

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

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 $x = 0$.