We have two independent observations $X_1$ and $X_2$ both from a poission distribution with $$X_1 \sim Pois(a\lambda) \\ X_2 \sim Pois(b\lambda)$$ with $a$ and $b$ known and positive.
We calculate the MLE and find that $\lambda = \frac{2n}{a\sum_{i=1}^nx_1+b\sum_{i=1}^nx_2}$
Calculated by the following:
$$L(a\lambda, b\lambda;x) = \prod_{i=1}^n f(x_i;a\lambda,b\lambda)=\prod_{i=1}^{n}ab\lambda^2e^{-\lambda(ax_1+bx_2)} $$ By taking the logarithm and calculating the first derivative with respect to $\lambda$ and setting everything to zero, we get the following: $$ \hat{\lambda} = \frac{2n}{a\sum_{i=1}^nx_{1i}+b\sum_{i=1}^nx_{2i}}$$
$$\text{For }n=2,\hat{\lambda}=\frac{2}{ax_1+bx_2}$$
Now, the question wants me to compare the estimator of lambda to the following sampling distribution estimators:
$$T_1 = \frac{X_1-X_2}{a-b}\\ T_2 = \frac{1}{2}\left(\frac{X_1}{a}+\frac{X_2}{b} \right)$$
To recommend an estimator for $\lambda$.
By taking the variance:
$$Var(T_1)=\frac{1}{(b-a)^2}\left(Var(X_1)-Var(X_2) \right) = \frac{1}{(b-a)^2}(\lambda-\lambda)=0 \\ Var(T_2) = \frac{1}{4}\left(\frac{1}{a^2}Var(X_1)+\frac{1}{b^2}Var(X_2)\right)=\frac{1}{4}\left(\frac{\lambda}{a^2}+\frac{\lambda}{b^2}\right)$$
How do I proceed from here?