You have two processes of measuring the air pollution, $X$ and $Y$. Both processes deliver values which are normal distributed around $\mu$: $X ~ N(\mu, \sigma_x^2)$ and $Y ~ N(\mu, \sigma_y^2)$.
I have two estimation processes assuming that $\sigma_x^2 < \sigma_y^2$:
$T_1 = \frac{X_1 + X_2}{2}$ and $T_2 = \frac{X_1+Y_1+Y_2}{3}$.
Show that $E(T_1) = \mu$ and calculate $Var(T_2)$.
This was not a problem at all, I did it this way: $E(T_1) = 0.5 * (E(X_1)+E(X_2)) = 0.5 * (2\mu) = \mu$.
$Var(T_2) = (\frac{1}{3})^2 * (\sigma_x^2 + 2 * \sigma_y^2) $. Can this be correct?
Under which conditions for the relation $\sigma_y^2/\sigma_x^2$ is $T_2$ a more effective estimation than $T_1$?
I know that an estimation $S1$ is more effictive that an estimaton $S2$ if $E((S2 - \frac{\sigma_Y^2}{\sigma_X^2})^2) < E((S1 - \frac{\sigma_Y^2}{\sigma_X^2})^2)$, but I don't know how to use this information...
I am thankful for any tips!
Your computation of $Var(T_2)$ is correct, and similarly, $Var(T_1) = \frac{\sigma_x^2}{2}$.
$T_2$ is more effective than $T_1$ when $Var(T_2) < Var(T_1)$. This requires: $$\frac{1}{9}*(\sigma_x^2 + 2*\sigma_y^2) < \frac{\sigma_x^2}{2}$$ $$ \frac{\sigma_y^2}{\sigma_x^2} < \frac{7}{4}$$