Question

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?

Answer 1

`\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)`