We have $N_1$, $N_2$, $N_3$ normally distributed random variables with three different means $µ_i$, but all with the same standard deviation $\sigma=1$.

I want to know the probability that $N_1$ is the largest of the three random variables.

I can solve this for two random variables. In this case I create a new random variable $d_{1,2}$ with $d_{1,2} = N_1 - N_2$. $d_{1,2}$ is normally distributed with mean $µ_1 - µ_2$ and standard deviation $\sqrt 2$. It is easy to calculate the portion of the distribution of $d_{1,2}$ that is larger than $0$. And if $d_{1,2} > 0$, this is equivalent to $N_1 > N_2$. Example: $µ_1 = \sqrt 2$, $µ_2 = 0$, $p(d_{1,2} > 0) = \Phi(1) = 0.84$.

I have not yet found a solution for three normally distributed random variables.


You have to specify the joint distribution. I gather from your question that you are additionally given that the three normal random variables are independent.

If so, the question can be framed as - what is the probability that $N_1 > N_2 \wedge N_3$. The answer to that question is a simply integral of the joint density function $f(N_1, N_2, N_3)=f_1(N_1)f_2(N_2) f_3(N_3)$ over a certain region.

$$\int_{N_1=N_2}^{\infty}\int_{N_2=N_3}^{\infty}\int_{N_3=-\infty}^{\infty} f_1f_2f_3 dN_3dN_2dN_1 +\int_{N_1=N_3}^{\infty}\int_{N_2=-\infty}^{N_3}\int_{N_3=-\infty}^{\infty} f_1f_2f_3 dN_3dN_2dN_1$$ where the expressions for the integrands are known.

  • $\begingroup$ This would be true for two distributions as well. However, the cited solution for two dimensions is ... easier (don't know how to say it correctly). I have no need to integrate but can rely on the cumulative normal distribution (which is a result of integration, I know, but which is simply available on nearly any math software, Matlab: normcdf). Is there no such solution for 3 dimensions? $\endgroup$
    – Christian
    Jan 26 '18 at 15:14
  • $\begingroup$ @Christian Yes, I realized that you were trying to get that other type of answer. But it is not possible in 3 dimensions (the boundary of the region to integrate over isn't linear anymore because of the max function). Once the regions to integrate over are identified correctly, it's not a terrible integration. $\endgroup$ Jan 26 '18 at 15:35
  • $\begingroup$ Thanks for answering my comment. "not a terrible integration"... I was considering to do it inside an optimization loop, so cpu time matters. So you are sure that there is no "that other type of answer" for 3 dimensions.... bad luck. But thanks anyway $\endgroup$
    – Christian
    Jan 28 '18 at 8:01

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