Help needed in the computation of the variance of this random variable

Question

Suppose we have random variables $X_1,\cdots,X_m$ bernoulli distributed with probability $p_u$, $D_1,\cdots,D_m \sim Exp(\lambda_d)$, $P_1,\cdots,P_m \sim Exp(\lambda_P)$ and let $d_i := X_i D_i + (1-X_i)P_i$ for $i=1,\cdots,m$. I have computed that $E(d_i) = p_u 1/\lambda_d + (1-p_u) 1/\lambda_P$. Now I want to compute the variance: I get $Var(d_i) = 1/\lambda_d^2(2p_u-p_u^2)+1/\lambda_P^2(1-p_u^2)$. But somehow I do not see how this result is correct, since it is not symmetric in $p_u$ and $1-p_u$. Could somebody please provide the correct answer?

Answer 1

$$Var(d_i) = Var(X_iD_i) + Var((1-X_i)P_i) + 2Cov(X_iD_i, (1-X_i)P_i)$$

$$Var(X_iD_i) = Var(E(X_iD_i|X_i)) + E(Var(X_iD_i|X_i)) = Var\left(\frac{X_i}{\lambda_D}\right) + E\left(\frac{X_i^2}{\lambda_D^2}\right)$$

$$\implies Var(X_iD_i) = \frac{p_u(2-p_u)}{\lambda_D^2}$$

Similarly,

$$Var((1-X_i)P_i) = \frac{(1-p_u)(1+p_u)}{\lambda_P^2}$$

$$Cov(X_iD_i, (1-X_i)P_i) = E(X_iD_i(1-X_i)P_i) - E(X_iD_i)E((1-X_i)P_i)$$ $$E(X_iD_i(1-X_i)P_i) = E(D_i)E(P_i)E(X_i(1-X_i))$$ $$E(X_iD_i)E((1-X_i)P_i) = E(D_i)E(P_i)E(X_i)E(1-X_i)$$

$$\implies Cov(X_iD_i, (1-X_i)P_i) = E(D_i)E(P_i)(E(X_i(1-X_i)) - E(X_i)E(1-X_i)) $$ $$\implies Cov(X_iD_i, (1-X_i)P_i) = -E(D_i)E(P_i)Var(X_i) = -\frac{p_u(1-p_u)}{\lambda_D\lambda_P}$$

$$Var(d_i) = \frac{p_u(2-p_u)}{\lambda_D^2} + \frac{(1-p_u)(1+p_u)}{\lambda_P^2} -2\frac{p_u(1-p_u)}{\lambda_D\lambda_P}$$