If the following is a probability distribution function: #f(x)=k(e^(-x^2)+e^-x)#, what is #k# and what is the variance?
1 Answer
Mar 31, 2016
There is no such
Explanation:
The area under a probability density function is 1.
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
That is because the integral
#int_{-oo}^{oo} e^{-x} "d"x #
itself is divergent.
Therefore, there is no solution for
