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

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Answer 1 · 2016-03-31T02:21:46

There is no such $k$ .

Therefore,

$int_{-oo}^{oo} f(x) "d"x = 1$

or

$int_{-oo}^{oo} k(e^{-x^2}+e^{-x}) "d"x = 1$

However, the integral

$int_{-oo}^{oo} k(e^{-x^2}+e^{-x}) "d"x$

does not converge for any value of $k$ .

That is because the integral

$int_{-oo}^{oo} e^{-x} "d"x$

itself is divergent.

Therefore, there is no solution for $k$ .

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