A random variable $X$ has mean $\mu$ and variance $2$ and an independent random variable $Y$ has mean $3\mu$ and variance $7$. Find the values of $a$ and $b$ if $aX+bY$ has minimum variance. The values obtained in a single observation of each of $X$ and $Y$ are $10$ & $15$ respectively. Obtain a best estimator for $\mu$ and explain in what it is best.
This is my attempt. In this question by minimum variance is it meant that it is $=0$. Then $a=0$ and $b=0$. And since there is only one observation is used there is no variability. Hence the estimator used should be unbiased. That is: $$ E(aX+bY)=\mu$$
$$a\mu+3b\mu=\mu $$ When $a+3b=1$ we get the most efficient estimator. Therefore it is the best estimator. Is my approach correct? Any help please.
I suspect you want a minimum variance unbiased estimator. So $a$ and $b$ cannot be chosen to be $0$. In order for $aX+bY$ to have mean $\mu$, that is, to make the estimator unbiased, we will need $a\mu+3b\mu=\mu$, so $a+3b=1$.
The variance of $aX+bY$ is $2a^2+7b^2$. This is what we want to minimize, subject to the condition $a+3b=1$. So $a=1-3b$, and our variance, in terms of $b$, is $$2(1-3b)^2+7b^2=25b^2-12b+2.$$ Use the usual calculus (or completing the square) procedure to find the best $b$. It turns out to be $\frac{6}{25}$.