# 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?

Nov 17, 2016

$\setminus \mu = \left(56 + \setminus \log 9\right) k$
$\setminus \sigma = \setminus \sqrt{\left(\frac{1160}{3} + \setminus \log 9\right) k - {\left(56 + \setminus \log 9\right)}^{2} {k}^{2}}$

#### Explanation:

First, let's rewrite the PDF (probability density function) as follows:
$p \left(x\right) = k \left(1 + \frac{1}{x - 1}\right)$ for $x \setminus \in \left[2 , 10\right]$
and
$p \left(x\right) = 0$ for $x \setminus \notin \left[2 , 10\right]$
(the PDF is always 0 outside of the range of $x$)

We will use the formula for the ${n}^{\text{th}}$ moment of a random variable $X$:
$\setminus m a t h \boldsymbol{E} \left[{X}^{n}\right] = \setminus {\int}_{- \setminus \infty}^{+ \infty} {x}^{n} p \left(x\right) \text{d} x$

The mean is the first moment (simply put $n = 1$):
$\setminus \mu = \setminus m a t h \boldsymbol{E} \left[X\right] = \setminus {\int}_{- \setminus \infty}^{+ \infty} x p \left(x\right) \text{d"x=\int_2^10 kx(1+1/(x-1))"d} x$
(because outside $\left[2 , 10\right]$ the PDF is $0$ and $\setminus \int 0 \text{d} x = 0$)
$\setminus \mu = k \setminus {\int}_{2}^{10} \left(x + \frac{x}{x - 1}\right) \text{d} x = \left(56 + \setminus \log 9\right) k$

Now, to find the standard deviation, we'll first evaluate the variance form the following formula:
$\setminus {\sigma}^{2} = \setminus m a t h \boldsymbol{E} \left[{X}^{2}\right] - \setminus m a t h {\boldsymbol{E}}^{2} \left[X\right]$

$\setminus m a t h \boldsymbol{E} \left[{X}^{2}\right] = \setminus {\int}_{2}^{10} k {x}^{2} \left(1 + \frac{1}{x - 1}\right) \text{d} x = \left(\frac{1160}{3} + \setminus \log 9\right) k$
$\setminus {\sigma}^{2} = \left(\frac{1160}{3} + \setminus \log 9\right) k - {\left(56 + \setminus \log 9\right)}^{2} {k}^{2}$
$\setminus \sigma = \setminus \sqrt{\setminus {\sigma}^{2}} = \setminus \sqrt{\left(\frac{1160}{3} + \setminus \log 9\right) k - {\left(56 + \setminus \log 9\right)}^{2} {k}^{2}}$

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 $p \left(x\right)$ isn't the PDF of the variable $X$).

$1 = \setminus {\int}_{- \setminus \infty}^{+ \setminus \infty} p \left(x\right) \text{d"x=k\int_2^10(1+1/(x-1))"d} x = k \left(8 + \setminus \log 9\right)$
So:
$k = \frac{1}{8 + \setminus \log 9}$