Question

What are the mean and standard deviation of the probability density function given by `p(x)=ke^-x` for `x in [0,1]`, in terms of k, with k being a constant such that the cumulative density across all x is equal to 1?

Answer 1

Mean: `mu = k (1-2/e)`
Std Dev: `sigma= sqrt(k (2-5/e) - (k (1-2/e) )^2)`

Explanation:

Let `X` be some random variable with the probability density function `p` Integration by parts is needed but I have just quoted the integral result:

The mean is given by;

`E(X) = int_D \ xp(x) \ dx`
`\ \ \ \ \ \ \ \ \ = int_0^1 xke^(-x) \ dx`
`\ \ \ \ \ \ \ \ \ = k \ int_0^1 xe^(-x) \ dx`
`\ \ \ \ \ \ \ \ \ = k [-(x+1)e^(-x)]_0^1`
`\ \ \ \ \ \ \ \ \ = -k [(x+1)e^(-x)]_0^1`
`\ \ \ \ \ \ \ \ \ = -k (2e^(-1)-e^0)`
`\ \ \ \ \ \ \ \ \ = k (1-2/e)`

And, the Variance is given by

`Var(X)=E(X^2)-E^2(X)`

We can calculate, `E(X^2)`, again integration by parts is required bu the result is just quoted:

`E(X^2) = int_D \ x^2p(x) \ dx`
`\ \ \ \ \ \ \ \ \ \ \ = k \ int_0^1 x^2e^(-x) \ dx`
`\ \ \ \ \ \ \ \ \ \ \ = k [-(x^2+2x+2)e^(-x) ]_0^1`
`\ \ \ \ \ \ \ \ \ \ \ = -k [(x^2+2x+2)e^(-x) ]_0^1`
`\ \ \ \ \ \ \ \ \ \ \ = -k (5e^(-1) -2e^0]`
`\ \ \ \ \ \ \ \ \ \ \ = k (2-5/e)`

And so the standard deviation is given by

`sigma^2= k (2-5/e) - (k (1-2/e) )^2`

Incidental, As `p(x)` represents a probability density function over `[0,1]` then:

`int_0^1 \ p(x) \ dx =1`
`:. int_0^1 \ ke^(-x) \ dx =1`
`:. [(ke^(-x))/(-1)]_0^1 =1`
`:. -k [(e^(-x))]_0^1 =1`
`:. -k (1/e-e^0) =1`
`:. k (1-1/e) =1`
`:. k =1/(1-1/e)`
`:. k =e/(e-1)`

And so the probability density function is

`p(x) = e/(e-1) e^(-x)`