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

Apr 12, 2018

$\left\langle x\right\rangle = \frac{n}{n + 2}$

$\sigma = \frac{1}{2 + n} \sqrt{\frac{2 n}{\left(3 + n\right)}}$

#### Explanation:

Find normalisation constant $k$ so that ${\int}_{m a t h \boldsymbol{R}} \mathrm{dx} \setminus m a t h \boldsymbol{P} \left(x\right) = 1$:

$k {\int}_{0}^{1} \setminus \mathrm{dx} \setminus {x}^{n} - {x}^{n - 1} = k {\left(\frac{{x}^{n + 1}}{n + 1} - \frac{{x}^{n}}{n}\right)}_{0}^{1} = - k \frac{1}{n \left(n + 1\right)}$

$\implies k = - n \left(n + 1\right)$

Mean

$\left\langle x\right\rangle = k {\int}_{0}^{1} \setminus \mathrm{dx} \setminus x m a t h \boldsymbol{P} \left(x\right)$

$= k {\int}_{0}^{1} \setminus \mathrm{dx} \setminus {x}^{n + 1} - {x}^{n} = k {\left(\frac{{x}^{n + 2}}{n + 2} - \frac{{x}^{n + 1}}{n + 1}\right)}_{0}^{1}$

$= \frac{n}{n + 2}$

Standard deviation

Using the relation: ${\sigma}^{2} = \left\langle {x}^{2}\right\rangle - {\left\langle x\right\rangle}^{2}$

$\left\langle {x}^{2}\right\rangle = {\int}_{0}^{1} \setminus \mathrm{dx} \setminus {x}^{2} m a t h \boldsymbol{P} \left(x\right)$

$= k {\int}_{0}^{1} \setminus \mathrm{dx} \setminus {x}^{n + 2} - {x}^{n + 1} = k {\left(\frac{{x}^{n + 3}}{n + 3} - \frac{{x}^{n + 2}}{n + 2}\right)}_{0}^{1}$

$= \frac{n \left(n + 1\right)}{\left(n + 2\right) \left(n + 3\right)}$

Therefore:

$\sigma = \sqrt{\frac{n \left(n + 1\right)}{\left(n + 2\right) \left(n + 3\right)} - {\left(\frac{n}{n + 2}\right)}^{2}}$

$= \frac{1}{2 + n} \sqrt{\frac{2 n}{\left(3 + n\right)}}$