I have a survival function with multiple terms like:
$$ S(X) = {1 \over 3}e^{(-x/3)}+{2 \over 3}e^{(-x/6)}$$
I want to calculate the expected value and variance.
$$ E(X) = \int_0^{\infty} S(X) = 5 $$
For the variance I used:
$$ S(X) = 1 - F(X) = 1 - ({1 \over 3}e^{(-x/3)}+{2 \over 3}e^{(-x/6)}) $$
$$ f(x) = {1 \over 9 } e^{-x/3} + {2 \over 18 } e^{-x/6} $$
$$ \sigma^2 = 2 \int_0^{\infty} xS(X) dx - E(X)^2 = \int_0^{\infty} x^2f(x)dx -E(X)^2 = 54 - 25 = 29$$
For exponential distributions, the mean is $ 1 \over \lambda$ and the variance $1 \over \lambda^2 $, so that the $ E(X) = \sqrt{\sigma^2} = \sigma$. This is not the case for the formula above and somehow I think I understood the whole thing not completely.
The first question is, is my variance correct? If yes, how can I imagine it visually? I always thought about it as from 0 to $\sigma^2$ with the mean in the middle.
Thank you, WiPu
Your problem concerns a random sampling of one of two exponential distributions. So$$\Bbb EX=\tfrac133+\tfrac236=5,\,\Bbb EX^2=\frac132\cdot3^2+\frac232\cdot6^2=54,$$as per your calculations. More generally, if $X$ samples $\operatorname{Exp}(\lambda_i)$ with probability $p_i$ then$$\Bbb EX=\sum_i\frac{p_i}{\lambda_i},\,\Bbb EX^2=\sum_i\frac{2p_i}{\lambda_i^2}=\sum_{ij}\frac{2p_ip_j}{\lambda_i^2}\implies\sigma^2=\sum_{ij}\left(\frac{2p_ip_j}{\lambda_i^2}-\frac{p_ip_j}{\lambda_i\lambda_j}\right).$$