Suppose $X_i$ ~ Poisson$(λ_i ), i=1,2,3$. The $X_i$'s are independent.
Let $Y_1 = X_1 + X_2$ and $Y_2 = X_2 + X_3$. Find the joint MGF and the covariance using the properties of expectation.
Here I know the individual MGF's: $$m_{Y_1}(t_1)=e^{(\lambda_1+\lambda_2)(e^{(t_1-1)})}$$ $$m_{Y_2}(t_2)=e^{(\lambda_2+\lambda_3)(e^{(t_2-1)})}$$
I ended up multiplying these to get the joint MGF but then realized that was probably wrong since the $Y$'s both depend on $X_2$? How should I do this?
EDIT: I have done what was suggested and get the following: $$m_{Y_1Y_2}(t_1,t_2)=^{\lambda_1(e^{t_1}-1)}e^{\lambda_2(e^{t_1+t_2}-1)}e^{\lambda_3(e^{t_2}-1)}$$
For the covariance, do I just do this as: $$Cov(Y_1,Y_2)=E(Y_1Y_2)-E(Y_1)E(Y_2)$$ $$=E((X_1+X_2)(X_2+X_3))-E(X_1+X_2)E(X_2+X_3)$$ Would this mean I get a covariance of 0?
$$m_{Y_1,Y_2}(t_1,t_2)$$ $$=Ee^{t_1Y_1+t_2Y_2}$$ $$=Ee^{t_1X_1+(t_1+t_2)X_2+t_2X_3}$$ $$=Ee^{t_1X_1}Ee^{(t_1+t_2)X_2}e^{t_2X_3}$$ $$=m_{X_1}(t_1)m_{X_2}(t_1+t_2)m_{X_3}(t_2).$$