I have
$X \sim N(1,3), Y \sim N(1,2) \mbox{ independent}, \\ Z = X - Y - 1$
a)
$\begin{array}{lcl}\text{E}(Z^2) & = & \text{E}[(X-Y-1)^2] \\ & = & \text{E}(X^2+Y^2 +1 -2XY -2X +2Y) \\ & = & \text{E}(X^2) + \text{E}(Y^2) + \text{E}(1) + \text{E}(-2XY) + \text{E}(-2X) + \text{E}(2Y) \\ & = & \text{Var}(X) + \text{E}^2(X) + \text{Var}(Y) + \text{E}^2(Y) + 1 - 2\text{E}(XY) -2\text{E}(X) +2\text{E}(Y) \\ & = & 3 + (1)^2 + 2 + (1)^2 + 1 - 2\text{E}(X)\text{E}(Y) - 2(1)+2(1) \\ & = & 3+1+2+1+1-2(1)(1)-2+2 \\ & = & 6 \end{array}$
b)
$\begin{array}{lcl} \text{Var}(Z-X) & = & \text{Var}(Z) + \text{Var}(-X) + 2\text{Cov}(Z,-X) \\ & = & \text{Var}(Z) + \text{Var}(-X) + 2(-3) \\ & = & \text{Var}(X-Y-1) + (-1)^2\text{Var}(X) - 6 \\ & = & \text{Var}(X) + \text{Var}(-Y) +\text{Var}(-1) + \text{Var}(X) - 6 \\ & = & 3 + (-1)^2\text{Var}(Y) + 0 + 3 - 6 \\ & = & 3+2+3-6 \\ & = & 2 \end{array}$
where the covariance is:
$\begin{array}{lcl} \text{Cov}(Z, -X) & = & \text{E}\left \{[Z-\text{E}(Z)][-X-\text{E}(-X)] \right \} \\ & = & \text{E} \left \{ [X-Y-1-\text{E}(X-Y-1)][-X+\text{E}(X)] \right \} \\ & = & \text{E} \left \{ [X-Y-1-(\text{E}(X)+\text{E}(-Y)+\text{E}(-1))][-X+1] \right \} \\ & = & \text{E} \left \{ [X-Y-1-\text{E}(X)+\text{E}(Y)+1][-X+1] \right \} \\ & = & \text{E}[(X-Y-1-1+1+1)(-X+1)] \\ & = & \text{E}(-X^2 + X + XY - Y) \\ & = & \text{E}(-X^2) + \text{E}(X) + \text{E}(XY) + \text{E}(-Y^) \\ & = & -\text{E}(X^2) + \text{E}(X) + \text{E}(XY) - \text{E}(Y) \\ & = & -(\text{Var}(X) + \text{E}^2(X)) + 1 + \text{E}(X)\text{E}(Y) - 1 \\ & = & -3-(1)^2 +1 + 1(1) - 1 = -3 \end{array}$
c)
$\begin{array}{lcl} P(Z \le 1) & = & P(X-Y-1 \le 1) \\ & = & P (X-Y \le 0)\end{array}$
$X-Y \sim N(1-1, 3+2) = N(0,5) \\ \mbox{let } T = X-Y \mbox{ then }$
$\begin{array}{lcl} P (T \le 0) & = & P \left (\frac{T-0}{\sqrt{5}} \le \frac{0-0}{\sqrt{5}} \right ) \\ & = & P(W \le 0) \\ & = & \phi(0) \\ & = & 0.50 \end{array}$
where $W \sim N(0,1)$
please, can you tell me if I have proceeded right? Many thanks!