# If the following is a probability distribution function: f(x)=k(e^(-x^2)+e^-x), what is k and what is the variance?

Mar 31, 2016

There is no such $k$.

#### Explanation:

The area under a probability density function is 1.

Therefore,

${\int}_{- \infty}^{\infty} f \left(x\right) \text{d} x = 1$

or

${\int}_{- \infty}^{\infty} k \left({e}^{- {x}^{2}} + {e}^{- x}\right) \text{d} x = 1$

However, the integral

${\int}_{- \infty}^{\infty} k \left({e}^{- {x}^{2}} + {e}^{- x}\right) \text{d} x$

does not converge for any value of $k$.

That is because the integral

${\int}_{- \infty}^{\infty} {e}^{- x} \text{d} x$

itself is divergent.

Therefore, there is no solution for $k$.