Suppose that the moment generating function of $(X, Y)$ is $ψ_{X,Y} (t, u) = exp(2t + 3u + t^2 + atu + 2u^2)$. Assume $(X,Y)$ is normal. Determine $a$ so that $X + 2Y$ and $2X − Y$ become independent.
Covariance matrix is $\lambda_{11}=1,\lambda_{12}=a,\lambda_{21}=a,\lambda_{22}=2$
$Cov(x+2y,2x-y)=2cov(x,x)-cov(x,y)+2cov(y,x)-2cov(y,y)=2-a+2a-4$ so $a=2$
but the answer is $4/3$. Can you explain?
If we know that $(X,Y)$ is jointly normal, then we can arrange the joint MGF of $(X,Y)$ in the form $$\psi_{X,Y}(t,u)=\exp\left(\mu_1t+\mu_2u+\frac{\sigma_1^2t^2+\sigma_2^2u^2+2\rho\sigma_1\sigma_2tu}{2}\right)$$
where $\mu_1=E(X),\mu_2=E(Y),\sigma_1^2=V(X),\sigma_2^2=V(Y)$ and $\rho=\text{Corr}(X,Y)$.
You have already done this part.
As $U=X+2Y$ and $V=2X-Y$ are linear combinations of two jointly normal variables $(X,Y)$, they are themselves jointly normal. So $(U,V)$ are independent iff $\text{Cov}(U,V)=0$.
Now, $\text{Cov}(U,V)=2V(X)-2V(Y)+3\text{Cov}(X,Y)=2(\sigma_1^2-\sigma_2^2)+3\rho\sigma_1\sigma_2$.
The rest is pure calculation.