Problem: Let $f_1(y)$ and $f_2(y)$ be density functions, and let $a$ be a constant such that $0\le a\le 1.$ Consider the function $f(y)=af_1(y)+(1-a)f_2(y).$
- Show that $f(y)$ is a density function. Such a density function is often referred to as a mixture of two density functions.
Suppose that $Y_1$ is a random variable with density function $f_1(y),$ and that $E(Y_1)=\mu_1$ and $\operatorname{Var}(Y_1)=\sigma_1^2;$ and similarly suppose that $Y_2$ is a random variable with density function $f_2(y),$ and that $E(Y_2)=\mu_2$ and $\operatorname{Var}(Y_2)=\sigma_2^2.$ Assume that $Y$ is a random variable whose density is a mixture of the densities corresponding to $Y_1$ and $Y_2.$
(i) Show that $E(Y)=a\mu_1+(1-a)\mu_2.$
(ii) Show that $\operatorname{Var}(Y)=a\sigma_1^2+(1-a)\sigma_2^2+a(1-a)[\mu_1-\mu_2]^2.$ [Hint: $E(Y_i^2)=\mu_i^2+\sigma_i^2,\;i=1,2.$]
My Work So Far:
Since $f_1$ and $f_2$ are non-negative, and $a$ and $1-a$ are also non-negative, it follows that $f$ is non-negative. We integrate: \begin{align*} \int_{\mathbb{R}}f(y)\,dy &=\int_{\mathbb{R}}[af_1(y)+(1-a)f_2(y)]\,dy\\ &=a\int_{\mathbb{R}}f_1(y)\,dy+(1-a)\int_{\mathbb{R}}f_2(y)\,dy\\ &=a+1-a\\ &=1, \end{align*} as required.
(i) We have \begin{align*} E(Y) &=\int_{\mathbb{R}}y\,f(y)\,dy\\ &=\int_{\mathbb{R}}y\,[af_1(y)+(1-a)f_2(y)]\,dy\\ &=a\int_{\mathbb{R}}y\,f_1(y)\,dy+(1-a)\int_{\mathbb{R}}y\,f_2(y)\,dy\\ &=aE(Y_1)+(1-a)E(Y_2), \end{align*} as expected.
(ii) We compute \begin{align*} \operatorname{Var}(Y) &=a^2\operatorname{Var}(Y_1)+(1-a)^2\operatorname{Var}(Y_2)+2a(1-a)\operatorname{Cov}(Y_1,Y_2)\\ &=a^2\sigma_1^2+(1-a)^2\sigma_2^2+2a(1-a)[E(Y_1Y_2)-\mu_1\mu_2]. \end{align*} We can arrive at this expression also by expanding out $V(Y)=E(Y^2)-(E(Y))^2.$ I see no way to simplify this. Indeed, if we assume $Y_1$ and $Y_2$ are independent, then the variances merely add, because the covariance disappears. It looks wrong to me, but I have verified that the identity holds for a couple of uniform distributions. I keep going around in circles on this one. Any ideas?
Many thanks for your time!
