Calculating expected value and variance of a probability density function
If I was given a probability density function:
$$f(y) = \left\{\begin{array}{ll}\frac{3y^2(4-y)}{64} & \textrm{for } 0 \leq y \leq 4\\ 0 & \textrm{elsewhere} \end{array}\right.$$
for expected value would that just be the following integral? $$\int_{0}^{4} yf(y)\,\textrm{d}y$$
I do not know how I would calculate the variance though. Any tips?
Thanks
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$\begingroup$You can either find the variance directly by applying the law of the unconscious statistician with $g(y)=(y-{\rm E}[Y])^2$, that is, $$ \mathrm{Var}(Y)={\rm E}[(Y-{\rm E}[Y])^2]=\int_0^4g(y)f(y)\,\mathrm dy, $$ or you could find ${\rm E}[Y^2]$ by the same formula with $g(y)=y^2$ and then use that $$ \mathrm{Var}(Y)={\rm E}[Y^2]-{\rm E}[Y]^2. $$
$\endgroup$ $\begingroup$For the expected value, you need to evaluate the integral $$\int_0^4 yf(y) dy =\int_0^4 {3y^3(4-y) \over 64} dy $$
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