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!

  • 5
    $\begingroup$ Presumably $0 \lt n \lt \infty$ and, in general, $\theta \gt 3$ (which does not have to be an integer). Otherwise the variance does not exist. You might want to compare this PDF to that of the F distribution: you will see that $n^2$ has an F distribution and that $y$ is a scale factor. The latter shows you may set $y=1$ and multiply the answer by $y^2$. That reduces the problem to finding the first two moments of the distribution with PDF $3n / (1+n^2)^{5/2}$. BTW, when you integrate it from $0$ to $\infty$ you had better get $1$ as the answer! $\endgroup$ – whuber Jan 14 '15 at 23:53

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.

  • $\begingroup$ You provide a very helpful and 101 intro to calculating the first two moments of a distribution. But, given that the OP does not know how to calculate a variance or a mean, do you think it is realistic to expect him to be able to compute the integrals required here, which are not exactly 101, unless we do impose $\theta = 4$? If the person asks: Q. How can I get to the moon? A. Build a space shuttle. If you have any problems, please let me know where you got stuck. $\endgroup$ – wolfies Jan 15 '15 at 17:02
  • $\begingroup$ @wolfies OP said he integrated his pdf to compute the mean, I don't see why he wouldn't be able to compute the variance. Often at introductory level, it's more difficult to lay out a mathematical problem than to resolve it. $\endgroup$ – mugen Jan 15 '15 at 17:09
  • $\begingroup$ @Raptors1102 if you can't work out the integrals, just show us where you got stuck and I or someone else will help you sort this out. $\endgroup$ – mugen Jan 15 '15 at 17:10
  • $\begingroup$ How do you obtain the equalities: $E[(X-\mu)^2] = \int_{-\infty}^{\infty}{(x-\mu)^2 f(x) dx}$ and $E[X^2] = \int_{-\infty}^{\infty}{x^2 f(x) dx}$ Can you point me to a proof of this, or to the property of integrals that is used to prove this? $\endgroup$ – John Smith Optional Jan 30 '19 at 15:10

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 $f(n)$:

enter image description here

The density is well-defined provided $\theta>1$. The mean $E_f[N]$ is:

enter image description here

and the variance of $N$ is:

enter image description here

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 $Var(N) = y^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.