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Consider three independent, normally distributed RVs: $YA \sim N(a,\sigma ^{2}),$ $% YB\sim N(b,\sigma ^{2})$ and $YC\sim N(c,\sigma ^{2})$.

What is the probability that $YA$ is the maximum?: $$\Pr (YA>YB \;\;\mbox{and} \;\;YA>YC)=IA=\int_{-\infty }^{\infty }\Phi (\frac{Y-b}{% \sigma })\Phi (\frac{Y-c}{\sigma })\phi (\frac{Y-a}{\sigma })dY$$

My first thought was to differentiate $IA$ with respect to $b$ and $c$, complete the square and exploit that $$\int_{-\infty }^{\infty }\frac{\sqrt{3}}{\sigma}\phi \left(\frac{Y-\frac{a+b+c}{2}}{% \sigma /\sqrt{3} }\right)dY=1$$ to remove the integral over $Y$. Therefore, $$\frac{d^2IA}{dbdc}\propto \exp \left(-\frac{a^2-2 a y+b^2-2 b y+c^2-2 cy+3 y^2}{2 \sigma ^2}\right)$$. Unfortunately, I cannot figure how to integrate with respect to $b$ and $c$ to get back to $IA$. Any suggestions would be much appreciated.

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    $\begingroup$ Are the variables independent? $\endgroup$ – Liviu Nicolaescu May 26 '15 at 15:19
  • $\begingroup$ Yes, YA, YB and YC are independent. $\endgroup$ – Alastair Smith May 26 '15 at 15:56
  • $\begingroup$ I have added the word independent in the statement of your question. $\endgroup$ – Liviu Nicolaescu May 26 '15 at 17:24
  • $\begingroup$ I don't believe your integral gives the correct answer. Even when $a=b=c=0$ it seems to be incorrect. $\endgroup$ – Greg Martin May 26 '15 at 23:39
  • $\begingroup$ Following on from @GregMartin, the first expression you write for $IA$ is missing a $1/\sigma$ factor (assuming $\phi$ is the density of $N(0,1)$). I haven't looked at the rest. $\endgroup$ – P.Windridge May 30 '15 at 17:29
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Since Google brought me here, I may as well share where I got solving this exact problem.

Note that:

$\Pr \left( YA>YB\text{ and }YA>YC\right) =\Pr \left( YB-YA<0\text{ and } YC-YA<0\right) $

and the distribution of $YB-YA\mid YA$ is simply $\phi _{\mu _{B}-YA,\sigma_{B}^{2}}$

The probability density of $YA=x$ and $YB-x=y$ and $YC-x=z$ is

$\Pr \left( YA=x\cap YB-x=y\cap YC-x=z\right) =\phi _{\mu _{B}-x,\sigma_{B}^{2}}\left( y\right) \phi _{\mu _{C}-x,\sigma _{C}^{2}}\left( z\right)\phi _{\mu _{A},\sigma _{A}^{2}}\left( x\right) $

Hence

$\Pr \left( YA>YB\text{ and }YA>YC\right) =\int_{-\infty }^{\infty }\left(\int_{-\infty }^{0}\phi _{\mu _{B}-x,\sigma _{B}^{2}}\left( y\right)dy\int_{-\infty }^{0}\phi _{\mu _{C}-x,\sigma _{C}^{2}}\left( z\right)dz\right) \phi _{\mu _{A},\sigma _{A}^{2}}\left( x\right) dx$

This expression can be simplified by convolution to get rid of the outer integral. But, like the cumulative normal distribution, there is no analytical solution.

If anyone wants the final answer, you can email me. But, also, if anyone knows a good algorithm for estimating this expression numerically, please email me.

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Since $\begin{pmatrix}YA-YB\\YA-YC\end{pmatrix}\stackrel{\text{dist}}{=} N\left(\begin{pmatrix}a-b\\a-c\end{pmatrix},\begin{pmatrix}2\sigma^2&\sigma^2\\ \sigma^2&2\sigma^2\end{pmatrix}\right)$, then \begin{align} \mathsf{P}((YA>YB)\cap(YA>YC))&=\mathsf{P}(YA-YB>0,YA-YC>0)\\ &=L\Bigl(\frac{b-a}{\sqrt2\sigma},\frac{c-a}{\sqrt2\sigma},\frac12\Bigr), \end{align} where $$L(h,k,\rho)=\frac{1}{2\pi\sqrt{1-\rho^2}}\int_h^\infty dx\int_k^\infty\exp\Bigl[ -\frac{x^2-2\rho xy+y^2}{2(1-\rho^2)}\Bigr]\,dy. $$ In the following book: S. Kotz et al., Continuous Multivariate Distributions, Vol. 1, 2nd Ed(2000), The Ch.46, Sec.4, p.264--. It gave a complete discussion about above bivariate normal integral--table and approximation.

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  • $\begingroup$ How is your answer different from @Nicolas' answer except for the additional reference, am I missing something? $\endgroup$ – Henry.L May 20 '17 at 11:35

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