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 $\theta_2$

  • 1
    $\begingroup$ Does your book (a) explain that the standard deviation is the square root of the variance and (b) provide some simple rules for computing variances of sums of random variables? (If it doesn't, then this question would be inappropriate for your book anyway.) $\endgroup$
    – whuber
    Dec 27, 2018 at 18:32
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    $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ Dec 27, 2018 at 18:35
  • $\begingroup$ @whuber the sum of two Norm. Dist. Rv is also normally distributed, and var(X+Y)=var(X)+var(Y), but how does this relate to standard error? Sorry if my questions sound stupid $\endgroup$
    – Lillys
    Dec 27, 2018 at 18:41
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    $\begingroup$ Can we always use the var(a+b) = var(a) + var(b) + 2 cov(a,b) formula when those three terms on the right hand side are estimates? $\endgroup$ Dec 27, 2018 at 18:59
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    $\begingroup$ I don't think this question is "stupid". Please add the [self-study] tag & add your understanding from your comments to the body of the question. $\endgroup$ Dec 27, 2018 at 19:36

1 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.


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