# Variance involving two independent variables

hello I have two independent variable P and Q. They are both non-negative. Now I define two new variables on them:

The first variable $$R_1=\alpha P+(1-\alpha)Q.$$ Since P and Q are independent, so $$Var(R_1)=\alpha^2 Var(P)+(1-\alpha)^2Var(Q)$$

The second variable R2 is a sort of compound variable: there is a probability of $$\alpha$$ that we get P and $$1-\alpha$$ probability to get Q. I work out the variance as $$Var(R_2)=\alpha E(P^2)+(1-\alpha)E(Q^2)-(\alpha E(P)+(1-\alpha)E(Q))^2$$

My intuition is that $$Var(R_2) \geq Var(R_1).$$ I wonder if anyone could help with the proof of the intuition above?

Here is an example. Let $$P=Q=(10,0.5;0,0.5)$$ which means there is 0.5 probability to get 10 and 0.5 probability to get 0. Let $$\alpha=0.5$$. Then $$R_1=(10,0.25;5,0.5;0,0.25)$$ and $$R_2=(10,0.5;0,0.5)$$. We get $$Var(R_1)=12.5$$ and $$Var(R_2)=25$$.

I tried a couple of other examples and they all show that $$Var(R_2)>Var(R_1)$$.

The inequality should follow from the Law of Total Variance. Also, I'm assuming $$\alpha \in [0,1]$$, as otherwise the construction doesn't make sense.

We need a slightly more formal definition of your compound variable. Let $$Z$$ be an independent, binary variable with $$P(Z=1)=\alpha$$, $$P(Z=0)=1-\alpha$$, and thus we can define $$R_2 := Z P + (1-Z)Q$$ (I assume this is what you mean).

We consider the variance of $$R_2$$ using the Law of Total Variance, conditioned on $$Z$$. I'll use subscripts on $$E$$ and $$Var$$ to denote the conditional expectation/variance.

$$Var(R_2) = E_Z[Var_Z(R_2)] + Var_Z(E_Z[R_2])$$

We want a sufficiently large lower bound on this variance, and we can discard the second term as $$\geq 0$$. Then, we can write out the first expectation for the two possible values $$Z=1$$ and $$Z=0$$. Recall, if we condition on $$Z=1$$, then $$R_2 = P$$.

$$Var(R_2) \geq E_Z[Var_Z(R_2)] = \alpha Var(P) + (1-\alpha)Var(Q)$$

Thus, our lower bound is a mixture of $$Var(P)$$ and $$Var(Q)$$. Compare this to your variance of $$R_1$$, and we note that for $$\alpha \in [0,1]$$, $$\alpha \geq \alpha^2$$, and $$(1-\alpha) \geq (1-\alpha)^2$$, which implies

$$Var(R_2) \geq Var(R_1).$$

While crude, this should be an easy way to understand the intuition behind the bound.
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