Standard error of $\widehat\theta_1 + \widehat\theta_2$

Question

I am working on some sample questions, and I came across one I have no clue how to answer.

How do Standard error of $\widehat\theta_1$, $\widehat\theta_2$, $\widehat\theta_1 + \widehat\theta_2$, relate to each other.

I know that the standard error is kind of the standard deviation of an estimator... I cant find any info in my book that relies to this question, or at least I can’t see the connection...

So i know that the estimator $\widehat\theta_i$ has a distribution and the stand. Dev of this distribution is called standard error. So the $se=\sqrt{var} $ of the distribuition of $\theta$

So I would answer the question like this

$$\sqrt{var(\widehat\theta_1 + \widehat\theta_2)}=\sqrt{var(\widehat\theta_1)+var(\widehat\theta_2)+ 2cov(\widehat\theta_1,\widehat\theta_2)}$$

Not sure if this is correct, and it does not tell me how it relates to se of $\theta_1$ and $\theta_2$

Answer 1

Some hints: Here are some special cases for you to consider. Suppose you have a random sample $X_1, \dots, X_8$ (so that $n=8),$ from a population with unknown mean $\mu$ (to be estimated) and known standard deviation $\sigma > 0.$

Two biased, independent estimators.

Let $\hat \mu_1 = (X_1 + X_2 + X_3 + X_4)/n,$ so that $E(\hat \mu_1) = \mu/2,$ variance $V(\hat \mu_1) = \frac{1}{64}(4\sigma^2) = \sigma^2/16,$ and $SE(\hat \mu_1) = \sigma/4.$

Notice that $E(\hat \mu_1) = \mu/2 \ne \mu,$ so that $\hat \mu_1$ is a seriously biased estimator of $\mu.$ But your Question says "estimator" not "good estimator."

Similarly, let $\hat \mu_2 = (X_5 + X_6 + X_7 + X_8)/n,$ so that $E(\hat \mu_2) = \mu/2,$ variance $V(\hat \mu_2) = \frac{1}{64}(4\sigma^2) = \sigma^2/16,$ and $SE(\hat \mu_2) = \sigma/4.$

Notice that the estimators $\hat \mu_1$ and $\hat \mu_2$ are independent because they use different elements of the random sample of size $n = 8.$ Then $$V(\hat \mu_1 + \hat \mu_2) = V(\hat \mu_1) + V(\hat \mu_2) = \sigma^2/8$$ and $SE(\hat \mu_1 + \hat \mu_2) = \sigma/\sqrt{8} > \sigma/4.$ Thus the SE of the sum is larger than the SE of the SE's of $\hat \mu_1$ and $\hat \mu_2.$

Nevertheless, $\bar X = \hat \mu_1 + \hat \mu_2$ is considered to be a better estimator of $\mu$ than either $\hat \mu_1$ or $\hat \mu_2,$ partly because it is unbiased, $E(\bar X) = \mu.$

Two estimators that are not independent.

Based on the same data, let $\hat \mu_3 = \bar X$ and $\hat \mu_4 = -\bar X.$ These estimators are obviously not independent. Also, for many distributions, $\bar \mu_4$ is an undesirable estimator, but an estimator nevertheless. I will leave the details of finding $SE(\hat \mu_3),\; SE(\hat \mu_4)$ and $SE(\hat \mu_3 + \hat \mu_4)$ to you. But you will find that that the third SE is smallest.