I suspect this is super-easy, but I haven't done any math in about ten years and I'm working with concepts that have been woefully explained...
I need to find the mean and median of a continuous random variable that has a probability density function of:
$f(x) = 2x^{-3}$ for $x > 1$
I know that this involves working out integrals and whatnot but, again, this is one of those concepts that wasn't actually explained to me.
$\endgroup$1 Answer
$\begingroup$let $X$ be some random variable with density $f$.
The mean of $X$ is $E(X)=\displaystyle\int_{-\infty}^{\infty}xf(x)dx=\int_{1}^{\infty}2x^{-2}dx=(-\frac2x|_{1}^{\infty}=2$.
And the median of $X$ is $M$ such that $\displaystyle\int_{-\infty}^{M}f(x)dx = \frac12$.
$\displaystyle\int_{-\infty}^{M}f(x)dx = \int_{1}^{M}2x^{-3}dx = (-x^{-2}|_{1}^{M} = 1-\frac1{M^2}$
therefore $\displaystyle\frac1{M^2}=\frac12$ which means $M=\sqrt2$
$\endgroup$ 3
