Variance of an estimator

Question

I was wondering how to compute the variance of the following estimator for the mean. Where $\{Y_1,Y_2,Y_3\}$ is a sample from an exponential distribution :
$$\hat{\theta} = \frac{Y_1 + Y_2}{4} + \frac{Y_3 + Y_1}{5}$$ I will assume that $\mathbb{E}(Y_i) = \frac{1}{\lambda} = \theta$ and similarly for the variance $\mathbb{V}(Y_i) = \frac{1}{\lambda^2} = \theta^2$

My reasoning is as follow :

1)We can re-write $\hat{\theta}$ as $\hat{\theta} = \frac{9Y_1 + 5Y_2 + 4Y_3}{20}$
2)$\mathbb{V}(\hat{\theta}) = \mathbb{V}(\frac{9Y_1 + 5Y_2 + 4Y_3}{20}) = \mathbb{V}(\frac{9}{20}Y_1) +\mathbb{V}(\frac{1}{4}Y_2)+\mathbb{V}(\frac{1}{5}Y_3) = \frac{9^2}{20^2}\mathbb{V}(Y_1) + \frac{1}{4^2}\mathbb{V}(Y_2) + \frac{1}{5^2}\mathbb{V}(Y_3) = \frac{9^2}{20^2}\theta^2 + \frac{1}{4^2}\theta^2 + \frac{1}{5^2}\theta^2 = \frac{122}{400}\theta^2$

I would have tought that we could also operate in the following manner :

1)Do NOT rewrite $\hat{\theta}$

2) $\mathbb{V}(\hat{\theta}) = \mathbb{V}(\frac{Y_1 + Y_2}{4}) +\mathbb{V}(\frac{Y_3 + Y_1}{5}) = \frac{1}{4^2}\mathbb{V}(Y_1 + Y_2) + \frac{1}{5^2}\mathbb{V}(Y_3 + Y_1) = \frac{2}{16}\theta^2 + \frac{2}{25}\theta^2 = \frac{50\theta^2 + 32\theta^2}{400} = \frac{82\theta^2}{400}$

I guess I'm missing something since the two results aren't the same. Which one is the correct one?And is this the correct approach to the general problem of computing the variance of an estimator?
Thanks!

Answer 1

Your first attempt is correct.

Your second attempt is incorrect because you ignored the covariance between $Z_1=Y_1+Y_2$ and $Z_2=Y_3+Y_1$.

Here is the correct way, accounting for the covariance:

$\hat\theta={1\over4}Z_1 +{1\over5}Z_2$

$V(\hat\theta)=\left({\partial\hat\theta\over \partial Z_1}\right)^2 V(Z_1) + \left({\partial\hat\theta\over \partial Z_2}\right)^2 V(Z_2) + 2\left({\partial\hat\theta\over \partial Z_1}{\partial\hat\theta\over \partial Z_2}\right) cov(Z_1,Z_2)$

$V(\hat\theta)={1\over16}2\theta^2+{1\over25}2\theta^2+2{1\over20}\theta^2={122\over400}\theta^2$

The covariance $cov(Z_1,Z_2)$ is easily worked out for this problem. Since it is the shared variance between $Z_1$ and $Z_2$, it is just the variance of $Y_1$.