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$$p(y) = \frac{c}{y^4}$$

I need to find the "mean & variance" of this exponential density function. Some pointers or thoughts that would explain would be most helpful.

How relatively likely is it that $Y$ occurs in an interval about $y=2$_dy_ compared to $y=3$_dy_?

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up vote 0 down vote accepted

This is not exponential, it is quadratic reciprocal. Anyway, do we know any boundaries for $y$?

This $p$ becomes a density function, if $\int_a^b p$ becomes $1$ (for the given interval $y\in [a,b]$). You can calculate this integral, and hence you find the $c$. [We perhaps may assume that $a=1$ and $b=+\infty$..]

The mean is defined as $E(Y) = \int_a^b y\cdot p(y)\ dy$, and the (square of) variance is $D^2(Y)= \int_a^b y^2\cdot p(y)\ dy - [E(Y)]^2$.

And to your last question: $p(y)$ is just telling the desired relative probabilistic 'for the infinitesimal', so the answer on that is $\frac{p(2)}{p(3)}$.

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