Suppose you were to toss 2 coins each with different probabilities of landing on heads 2 times each, one at a time. So coin 1 tossed twice. Then coin 2 tossed twice. Let N be the total number of heads. To calculate E[N], would you find E[number of heads on coin 1's tosses] and E[number of heads on coin 2's tosses] and sum them together? And to find the var[N] would it be the same concept?
[Math] Expected number of heads on coin flip
discrete mathematicsprobabilityrandom variables
Best Answer
For expectation - yes due to linearity of expectation. As for variance, you can do the same since the each toss (or each event) are independent. Note that it does not matter whether you toss each coin 'one at a time' or the order you do it in due to independence, of course.
Formally, if you let $X_1$ and $X_2$ be the indicator variables for each toss of the first coin ($1$ for heads, $0$ for tails) and the same with $Y_1$ and $Y_2$ for the tosses of the second coin, we have:
$$N=X_1+X_2+Y_1+Y_2$$
Then
$$E(N)=E(X_1+X_2+Y_1+Y_2)$$
$$=E(X_1)+E(X_2)+E(Y_1)+E(Y_2)$$
due to linearity of expectation.
Also, as $X_1,X_2,Y_1,Y_2$ are independent events, we have
$$Var(N)=Var(X_1+X_2+Y_1+Y_2)$$
$$=Var(X_1)+Var(X_2)+Var(Y_1)+Var(Y_2)$$
The general rules being used here are:
$$E(aX+bY)=aE(X)+bE(Y)\tag{linearity of expectation}$$
and if $X$ and $Y$ are independent,
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$