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?

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Answer 1 · 2018-04-12T08:10:45

$langle x rangle = n/(n+2)$

$sigma = 1/(2 + n) sqrt ((2 n)/( (3 + n)))$

Find normalisation constant $k$ so that $int_(mathbb R) dx \ mathbb P(x) = 1$ :

$k int_0^1 \ dx \ x^n - x^(n-1) = k ((x^(n+1))/(n+1) - (x^(n))/(n))_0^1 = - k 1/(n(n+1))$

$implies k = - n(n+1)$

Mean

$langle x rangle = k int_0^1 \ dx \ x mathbb P(x)$

$= k int_0^1\ dx \ x^(n+1) - x^n = k ((x^(n+2))/(n+2) - (x^(n+1))/(n+1))_0^1$

$= n/(n+2)$

Standard deviation

Using the relation: $sigma^2 = langle x^2 rangle - langle x rangle^2$

$langle x^2 rangle = int_0^1 \ dx \ x^2 mathbb P(x)$

$= k int_0^1 \ dx \ x^(n+2) - x^(n+1) = k ((x^(n+3))/(n+3) - (x^(n+2))/(n+2))_0^1$

$= (n(n+1))/((n+2)(n+3))$

Therefore:

$sigma = sqrt ( (n(n+1))/((n+2)(n+3)) - ( n/(n+2) )^2)$

$= 1/(2 + n) sqrt ((2 n)/( (3 + n)))$

What are the mean and standard deviation of the probability density function given by (p(x))/k=x^n-x^(n-1)(p(x))/k=x^n-x^(n-1) for x in [0,1] 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?