Question

What are the mean and standard deviation of the probability density function given by `p(x)=k-x+xe^(-x^2)` for `x in [0,4]`, 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 = -3.89`
`sigma = 5.2`

Explanation:

Given: `p(x)=k-x+xe^(-x^2)` for `x in [0,4]`

Required: Mean and Standard Deviation

Solution Strategy:
1) Normalize `p(x)`, i.e. determine the `k: P(x)=int_0^4 p(x) dx =1`
2) Calculate the mean, `mu = int_0^4 xp(x) dx`
3) Calculate the standard deviation, `sigma=int_0^4 (x-mu)^2p(x)dx`

1) `P(x)=int_0^4 [k-x+xe^(-x^2)]dx =1` Apply linearity and write
`1 = int_0^4 kdx -int_0^4xdx+int_0^4xe^(-x^2)dx`
`1 = kx -x^2/2+int_0^4xe^(-x^2)dx` Let's deal with the 3rd integral
`I= int_0^4xe^(-x^2)dx, " let " u = -x^2`
`I= -1/2int e^udu = -1/2 e^u` Undo substitution
`I= -1/2 [e^(-x^2)]_0^4 = 1/2-1/(2e^16)=(e^16-1)/(2e^16) -> 1/2`
`1 = 4K-8+1/2 => K=17/8`
Check: `17/2-16/2+1/2=1`

Thus `p(x)=17/8-x+xe^(-x^2)`

2) Mean: `mu= int_0^4 x[17/8-x+xe^(-x^2)]dx`
`mu=int_0^4 17/8xdx-int_0^4x^2dx+color(red)[int_0^4x^2e^(-x^2)dx]`
`= {17/16x^2-x^3/3 +color(red) [sqrt(pi)("erf"(x))/4-x/(2e^(x^2)]}}_0^4`

`~~ 17/16*16-64/3+color(red).443~~ -3.89`

3) `sigma^2= int_0^4 (x-3.89)^2 [17/2-x+xe^(-x^2)]dx`

`~~27.23134392285682`

`sigma = sqrt(27.23134392285682)~~ 5.2`