Consider a random vector (X, Y ) such that X ∼ Ber(0.6) and Y ∼ Ber(0.8). Is it possible to have Var[X + Y ] = 0.7?

Question

Consider a random vector (X, Y ) such that X ∼ Ber(0.6) and Y ∼ Ber(0.8). Is it possible to have Var[X + Y ] = 0.7? Thanks a lot.

Answer 1

\begin{align*} Var(X+Y) &= E[(X+Y)^2]-E[X+Y]^2\\ &= E[X^2] + 2E[XY] + E[Y^2] - (E[X]+E[Y])^2 \\ &= 0.6 + 2 E[XY] + 0.8 - (1.4)^2 \\ &= 2E[XY]-0.56 \\ &=0.7 \end{align*} which implies $0.63=E[XY]=P(X=1,Y=1)$. However, $P(X=1,Y=1) \le P(X=1)=0.6$, a contradiction.

Answer 2

You also tagged covariance, so if you really want to solve the problem by using the covariance, just calculate it through and use angryavian's idea.

\begin{align*} Var(X+Y) &= Var(X)+Var(Y)+2Cov(X,Y)\\ &= Var(X)+Var(Y)+2(E(XY)-E(X)E(Y)) \\ &\le Var(X)+Var(Y)+2(P(X=1)-E(X)E(Y)) \\ &=0.64 \end{align*}

Thus

$$Var(X+Y)\le 0.64<0.7.$$