Some time in the future, a lottery takes place, with winning number N. 3 of your friends from the future, John, Joe, and James, provide you with guesses on the number N.

John's guess a is randomly selected from a gaussian distribution centered at N with stdev x;

Joe's guess b is randomly selected from a gaussian distribution centered at N with stdev y;

James' guess c is randomly selected from a gaussian distribution centered at N with stdev z;

Given the values of a, x, b, y, c, z, what would be the best guess of N? Also define what "best" is.

  • 1
    $\begingroup$ You seem to be suggesting that lottery numbers do not need to be integers. Are you also assuming that your friends' errors are independent? $\endgroup$
    – Henry
    Commented Jan 1, 2020 at 2:11
  • $\begingroup$ I would expect that if you want to minimise the expected square of the error in your estimate of $N$, you may want a weighted average of $a$, $b$ and $c$; my intuitive guess is that the weights could be inversely proportional to the squares of the individual standard deviations $\endgroup$
    – Henry
    Commented Jan 1, 2020 at 2:48
  • $\begingroup$ @Henry their guesses are independent, and yes lottery numbers don't have to be integers. $\endgroup$
    – Joey Chen
    Commented Jan 1, 2020 at 3:12
  • $\begingroup$ @Henry what weighted average would you calculate, specifically, and why? $\endgroup$
    – Joey Chen
    Commented Jan 1, 2020 at 6:48
  • 1
    $\begingroup$ en.m.wikipedia.org/wiki/Inverse-variance_weighting $\endgroup$
    – seanv507
    Commented Jan 1, 2020 at 9:21

1 Answer 1


You can calculate and maximize the likelihood of N given a,b,c, with x,y,z being fixed.

The Likelihood of a value of N (the probability of sampling a,b,c given that the mean is N) is:

$LL_{a,b,c}(N) = Pr(a | x,N) \cdot Pr(b | y,N) \cdot Pr(c | z,N)$

With the distributions being independent and Gaussian, this is

$LL_{a,b,x}(N) = \frac{1}{x\sqrt{2\pi}} e^{-\frac{(a-N)^2}{2x^2}} \cdot \frac{1}{y\sqrt{2\pi}} e^{-\frac{(b-N)^2}{2y^2}} \cdot \frac{1}{z\sqrt{2\pi}} e^{-\frac{(c-N)^2}{2z^2}} = $ $\frac{1}{xyz(\sqrt{2\pi})^3} e^{-\frac{1}{2}(\frac{(a-N)^2}{x^2} +\frac{(b-N)^2}{y^2}+\frac{(c-N)^2}{z^2})}$

And we want to find the N that maximizes this likelihood. To find the maximum, we will search for a point where the derivative of the likelihood equals zero.

$\frac{d}{dN}LL_{a,b,c}(N) = \frac{1}{xyz(\sqrt{2\pi})^3}\cdot -\frac{1}{2}(\frac{2(a-N)}{x^2} + \frac{2(b-N) }{y^2}+ \frac{2(c-N)}{z^2}) e^{-\frac{1}{2}(\frac{(a-N)^2}{x^2} +\frac{(b-N)^2}{y^2}+\frac{(c-N)^2}{z^2})}$

This equals zero if and only if

$\frac{2(a-N)}{x^2} + \frac{2(b-N) }{y^2}+ \frac{2(c-N)}{z^2} = 0$

So we get that

$y^2z^2a - y^2z^2N + x^2 z^2 b- x^2 z^2 N + x^2 y^2 c-x^2 y^2 N = 0$

$N = \frac{y^2z^2a+ x^2z^2b + x^2y^2c}{y^2z^2 + x^2z^2 + x^2y^2}$

Is the maximum likelihood estimate.

  • $\begingroup$ Interestingly enough, @DaphnaKeidar and Henry, you guys have the same answer! While I do partially understand your proof, Daphna, I have not really taken a statistics course beyond the level of AP, so can only manage to piece together what you really mean with my slightly better calculus background. I was wondering if there was a more intuitive way of understanding it, like the weighted averages thing that Henri mentioned, but also that I do not quite understand. Thanks guys! $\endgroup$
    – Joey Chen
    Commented Jan 1, 2020 at 9:44
  • $\begingroup$ @Henry (couldn't mention you guys both at once) $\endgroup$
    – Joey Chen
    Commented Jan 1, 2020 at 9:45
  • $\begingroup$ Sorry, I did assume some knowledge in statistics.. I didn't use much beyond the likelihood function and Gaussian distribution pdf. An intuition would be that if the standard deviation is smaller, then the sample is probably closer to N. So you want to give more weight to the samples that have a smaller standard deviation. $\endgroup$ Commented Jan 1, 2020 at 9:56
  • $\begingroup$ Thanks @DaphnaKeidar ! I'm still not quite understanding what this specific weighted average is. Essentially, I don't quite know what this method of weighted averaging yields, and what the coefficients even mean. $\endgroup$
    – Joey Chen
    Commented Jan 1, 2020 at 10:00
  • $\begingroup$ The coefficients are derived using the standard deviations x,y,z $\endgroup$ Commented Jan 1, 2020 at 10:01

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