A random variable $n$ can be represented by its PDF
$$p(n) = \frac{(\theta - 1) y^{\theta-1} n}{ (n^2 + y^2)^{(\theta+1)/2}}.$$
$\theta$ is a positive integer and $y$ is a positive parameter. If $\theta=4$ how to you find the mean and variance?
My guess was to plug in $4$ of course and then integrate that function from $0$ to infinity. As for the variance I honestly have no clue. I have not taken statistics in a while so I admit I am a bit rusty. Any clues/help is appreciated. Thanks!
I'll give you a few hints that will allow you to compute the mean and variance from your pdf.
First of all, remember that the expected value of a univariate continuous random variable $E[X]$ is defined as $E[X] = \int_{-\infty}^{\infty}{x f(x) dx}$ as explained here, where the range of the integral corresponds to the sample space or support (say, $(-\infty, \infty)$ for a Gaussian distribution, $(0, \infty)$ for an exponential distribution).
Second, the mean of the random variable is simply it's expected value: $\mu = E[X] = \int_{-\infty}^{\infty}{x f(x) dx}$. It looks like you already covered that.
Third, the definition of the variance of a continuous random variable $Var(X)$ is $Var(X) = E[(X-\mu)^2] = \int_{-\infty}^{\infty}{(x-\mu)^2 f(x) dx}$, as detailed here. Again, you only need to solve for the integral in the support. Alternatively, it is sometimes easier to rely on the equivalent expression $Var(X) = E[(X-\mu)^2] = E[X^2] - (E[X])^2$, where the first term is $E[X^2] = \int_{-\infty}^{\infty}{x^2 f(x) dx}$ (see the definition of the expectation in the second paragraph) and the second term is $(E[X])^2 = \mu^2$.
Finally, you don't need to pick an arbitrary value for the parameter $\theta$ and plug it in the pdf. You can solve for the mean and the variance anyway. See, for example, mean and variance for a binomial (use summation instead of integrals for discrete random variables).
If you can't solve this after reading this, please edit your question showing us where you got stuck.
In case you get stuck computing the integrals referred to in the above post, here is an automated way to proceed. Given random variable $N$ has pdf $f(n)$:
The density is well-defined provided $\theta>1$. The mean $E_f[N]$ is:
and the variance of $N$ is:
where I am using the Expect and Var functions from the the mathStatica package for Mathematica to automate the nitty-gritties.
In the case of $\theta = 4$, the above results simplify to $E[N] = y$ and $Var(N) = y^2$.
