What are the mean and standard deviation of the probability density function given by #(p(x))/k=(1-1/(1-x))# for # x in [2,10]#, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

1 Answer
Nov 17, 2016

#\mu=(56+\log 9)k#
#\sigma=\sqrt{(1160/3+\log 9)k-(56+\log 9)^2k^2}#

Explanation:

First, let's rewrite the PDF (probability density function) as follows:
#p(x)=k(1+1/(x-1))# for #x\in [2,10]#
and
#p(x)=0# for #x\!in [2,10]#
(the PDF is always 0 outside of the range of #x#)

We will use the formula for the #n^"th"# moment of a random variable #X#:
# \mathbb{E}[X^n]=\int_{-\infty}^{+infty} x^n p(x)"d"x #

The mean is the first moment (simply put #n=1#):
#\mu=\mathbb{E}[X]=\int_{-\infty}^{+infty} xp(x)"d"x=\int_2^10 kx(1+1/(x-1))"d"x#
(because outside #[2,10]# the PDF is #0# and #\int 0 "d"x=0#)
#\mu=k\int_2^10 (x+x/(x-1))"d"x=(56+\log 9)k#

Now, to find the standard deviation, we'll first evaluate the variance form the following formula:
#\sigma^2=\mathbb{E}[X^2]-\mathbb{E}^2[X]#

#\mathbb{E}[X^2]=\int_2^10 kx^2(1+1/(x-1))"d"x=(1160/3+\log 9)k#
#\sigma^2=(1160/3+\log 9)k-(56+\log 9)^2k^2#
#\sigma=\sqrt{\sigma^2}=\sqrt{(1160/3+\log 9)k-(56+\log 9)^2k^2}#

This is how it looks in terms of #k#. If you want to find the exact values you can find #k# knowing that the integral of the PDF over the range of #x# is always equal to #1# (if it isn't #1# then the function #p(x)# isn't the PDF of the variable #X#).

#1=\int_(-\infty)^(+\infty) p(x)"d"x=k\int_2^10(1+1/(x-1))"d"x=k(8+\log 9)#
So:
#k=1/(8+\log 9)#