I have posted essentially the same question on reddit, but it has been some time now, and I have not yet received an answers, so I am also posting it here.

I have five measurements of speed $v_i\pm\sigma_i$, where i ranges from 1 to 5, and all the values of $v_i$ and $\sigma_i$ are different. The measurements are assumed to be independent of each other and normally distributed around a mean speed $v$. We can write a likelihood function for these measurements that has the following form:

$$L = \prod_1^5 \frac{1}{\sqrt{2\pi}\sigma_i}e^{-\frac{1}{2}(v_i-v')^2/(\sigma_i)^2}$$

There are two things that I am trying to figure out how to extract from this expression: The maximum likelihood estimator $v'$ for $v$, and the $\sigma(v')$ uncertainty associated with it. Now, it is fairly easy to extract the estimator $v$ from this: simply take the $\log(L)$, differentiate with respect to $v'$, and set the derivative to zero. Then solve for $v'$. You get something like $v' = \frac{\sum_1^5 \frac{v_i}{σ_i^2}}{\sum_1^5 \frac{1}{σ_i^2}}$.

But I am stuck with the standard deviation. The answers to this question say that the correct expression for $\sigma(v')$ is $\sigma(v')^2=\left(\sum_1^5 \frac{1}{\sigma_i^2}\right)^{-1}$ but for the love of me I could not figure out how this expression was obtained.

  • 1
    $\begingroup$ You already know the variances of the $v_i.$ So apply the bilinearity of variance to $v^\prime$ and use a tiny bit of algebra to simplify the result. $\endgroup$
    – whuber
    Dec 27, 2022 at 23:33
  • $\begingroup$ I'm sorry, I don't think I am familiar with the term 'bilinearity of variance,' could you provide a link on where I could read up on it? $\endgroup$
    – NX37B
    Dec 27, 2022 at 23:49
  • 1
    $\begingroup$ @NX37B I would suggest to look at the book, "Mathematical Statistics and Data Analysis", I believe it is also available for the public for free. In Section 8.5 of the book you will see the MLE equation written out for the mean and standard deviation parameter. $\endgroup$ Dec 28, 2022 at 3:33
  • $\begingroup$ Please search our site. $\endgroup$
    – whuber
    Dec 28, 2022 at 14:59


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