Question

Consider $k$ continuous-type distributions with the following characteristics: pdf $f_i(x)$, mean ${\mu }_i$, and variance ${\sigma }_i^2,i=1,2,\dots ,k.$ If $c_i\ge 0,i=1,2,\dots ,k$, and $c_1+c_2+\cdots +c_k=1$, show that the mean and the variance of the distribution having pdf $c_1{f}_1(x)+\cdots +c_k{f}_k(x)$ are $\mu =\sum _{i=1}^k{c}_i{\mu }_i$ and ${\sigma }^2=\sum _{i=1}^k{c}_i[{\sigma }_i^2+{({\mu }_i-\mu )}^2]$ respectively.

Short answer

The mean of the distribution is $\mu =\sum _{i=1}^k{c}_i{\mu }_i$ and the variance is ${\sigma }^2=\sum _{i=1}^k{c}_i({\sigma }_i^2+({\mu }_i-\mu {)}^2)$.

Step-by-step solution

For the mean

To calculate the mean, $\mu$, of the new distribution, simply take the expected value of the new distribution:$$\mu =E(c_1{f}_1(x)+\dots +c_k{f}_k(x))=c_{1}E(f_1(x))+\dots +c_{k}E(f_k(x))=c_1{\mu }_1+\dots +c_k{\mu }_k=\sum _{i=1}^k{c}_i{\mu }_i$$

For the variance

Now finding the variance, ${\sigma }^2$, is a bit more complicated. We can start by noting that:$$E(c_1{f}_1(x{)}^2+\dots +c_k{f}_k(x{)}^2)=c_{1}E(f_1(x{)}^2)+\dots +c_{k}E(f_k(x{)}^2)$$Then use the all squares identity: variance = expectation of square - square of expectation, we can write$${\sigma }^2=E[(c_1{f}_1(x)+\dots +c_k{f}_k(x){)}^2]-{\mu }^2$$Notice that $E[(c_1{f}_1(x)+\dots +c_k{f}_k(x){)}^2]$ can be expanded to yield both the variance terms and cross product terms. ${\mu }^2$ will be the square of the sum from step 1. When expanded, and terms composed, we will obtain:$${\sigma }^2=\sum _{i=1}^k{c}_i({\sigma }_i^2+({\mu }_i-\mu {)}^2)$$

Finishing up

The final results for $\mu$ and ${\sigma }^2$ are found by computing them for given values of $c_i$, ${\mu }_i$, and ${\sigma }_i$. Remember that all $c_i$ must equal to 1 when summed together, and each $c_i$ must be equal to or larger than zero.