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?

Question

1234 views around the world

You can reuse this answer
Creative Commons License

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)))#