What are the mean and standard deviation of the probability density function given by p(x)=k(x-x^2)  for  x in [0,1], in terms of k, with k being a constant such that the cumulative density across all x is equal to 1?

Nov 9, 2016

$E \left(X\right) = \frac{k}{12}$

$s d = \sqrt{\frac{3 k - 5 {k}^{2}}{60}}$

Explanation:

the mean $E \left(X\right)$ of a continuous pdf is given by

$E \left(X\right) = {\int}_{a l l x} x f \left(x\right) \mathrm{dx}$

so

$E \left(X\right) = {\int}_{0}^{1} k x \left(x - {x}^{2}\right) \mathrm{dx}$

$E \left(X\right) = k {\int}_{0}^{1} \left({x}^{2} - {x}^{3}\right) \mathrm{dx}$

$E \left(X\right) = k {\left[\frac{1}{3} {x}^{3} - \frac{1}{4} {x}^{4}\right]}_{\cancel{0}}^{1}$

$E \left(X\right) = k \left(\frac{1}{3} - \frac{1}{4}\right) = \frac{k}{12}$

$V a r \left(X\right) = E \left({X}^{2}\right) - {E}^{2} \left(X\right)$

$E \left({X}^{2}\right) = {\int}_{a l l x} {x}^{2} f \left(x\right) \mathrm{dx}$

$E \left({X}^{2}\right) = {\int}_{0}^{1} {x}^{2} k \left(x - {x}^{2}\right) \mathrm{dx}$

$E \left({X}^{2}\right) = k {\int}_{0}^{1} \left({x}^{3} - {x}^{4}\right) \mathrm{dx}$

$E \left({X}^{2}\right) = k {\left[\frac{1}{4} {x}^{4} - \frac{1}{5} {x}^{5}\right]}_{\cancel{0}}^{1}$

$E \left({X}^{2}\right) = k \left(\frac{1}{4} - \frac{1}{5}\right) = \frac{k}{20}$

$V a r \left(X\right) = E \left({X}^{2}\right) - {E}^{2} \left(X\right)$

gives

$V a r \left(X\right) = \frac{k}{20} - {\left(\frac{k}{12}\right)}^{2}$

$V a r \left(X\right) = \frac{3 k - 5 {k}^{2}}{60}$

$s d = \sqrt{\frac{3 k - 5 {k}^{2}}{60}}$