Problem:
A certain set of values if known to be normally distributed with $\sigma^2 = 1$. However, its mean is not known. The following three sample values are taken: $0, 1, 10$. We want the best possible estimate of the mean.
Answer:
First recall the general form for the density function for the normal distribution is: $$ N(x) = \dfrac{e^{-\dfrac{ (x-\mu)^2 }{2 \sigma^2}} }{ \sigma \sqrt{2 \pi } } $$ However, in this case, since $\sigma^2 = 1$ we have: $$ N(x) = \dfrac{e^{-\dfrac{ (x-\mu)^2 }{2 }} }{ \sqrt{2 \pi } } $$ The best estimate for the mean is a value $\mu$ such that the following expression is maximized: $$ N(0)N(1)N(10) $$ We define $f(\mu)$ to be: $$ f(\mu) = \left( e^{-\dfrac{ (0-\mu)^2 }{2 }} \right) \left( e^{-\dfrac{ (1-\mu)^2 }{2 }} \right) \left( e^{-\dfrac{ (10-\mu)^2 }{2 }} \right) $$ Now we simplify and then differentiate with respect to $\mu$. \begin{align*} f(\mu) &= \left( e^{-\dfrac{ (\mu)^2 }{2 }} \right) \left( e^{-\dfrac{ (\mu-1)^2 }{2 }} \right) \left( e^{-\dfrac{ (\mu-10)^2 }{2 }} \right) \\ % f( \mu ) &= e^{ \dfrac{ -\mu^2 - \mu^2 + 2\mu - 1 -\mu^2 + 20\mu - 100 }{2} } \\ f( \mu ) &= e^{\dfrac{ -3\mu^2 + 22\mu - 101 }{2}} \end{align*} Now to maximize $f( \mu )$ we maximize $g( \mu ) = -3\mu^2 + 22\mu - 101$. \begin{align*} g'( \mu ) &= -6\mu + 22 \\ -6\mu + 22 &= 0 \\ -3\mu + 11 &= 0 \\ \end{align*} Hence our best estimate for the population mean is: $$ \mu = \dfrac{ 11 }{3} $$ For comparison purposes, I want to compute $\bar{Y}$. I would expect my answer to be less than $\bar{Y}$. \begin{align*} \bar{Y} &= \dfrac{ 0 + 1 + 10 }{3} \\ \bar{Y} &= \dfrac{ 11}{3} \end{align*} Therefore, I am thinking my answer might be wrong.
