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I am looking for a closed-form version of this formula:

\begin{align*}\text{P} &= \int_{-\infty} ^{\infty} \left[1 - \left(\int_{-\infty} ^{t}\right.\right.\ldots\\ &\left.\left.\ldots\int_{-\infty} ^{t} \dfrac{1}{\sqrt{2\pi}\sigma_{2}} e^{-\frac{(x-\mu_{2})^{2}}{2\sigma_{2}^{2}}} \dfrac{1}{\sqrt{2\pi}\sigma_{3}} e^{-\frac{(x-\mu_{3})^{2}}{2\sigma_{3}^{2}}}\right.\right.\ldots\\ &\left.\left.\ldots \dfrac{1}{\sqrt{2\pi}\sigma_{M}} e^{-\frac{(x-\mu_{M})^{2}}{2\sigma_{M}^{2}}}~\text{d}x\ldots \text{d}x\right)\right] \dfrac{1}{\sqrt{2\pi}\sigma_{1}} e^{-\frac{(t-\mu_{1})^{2}}{2\sigma_{1}^{2}}}~\text{d}t \end{align*}

where the $\mu_i$'s and $\sigma_i^2$'s are the means and variances of the Gaussian distributions.

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    $\begingroup$ By using the Gaussian CDF you can immediately reduce this to a one-dimensional integral (is there any reason you did not do that?), but don't expect it to have any closed form except for very special choices of the parameters $\mu_i$ and $\sigma_i$. $\endgroup$
    – whuber
    Commented Feb 22, 2018 at 21:21
  • $\begingroup$ Thanks for your reply. I got P=\int_{-infinity}^{infinity} Q(...) e^(t-\mu_{1})^2/\sigma^2 dt, but not sure can I get a simpler form? $\endgroup$
    – Niki
    Commented Feb 22, 2018 at 21:33
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    $\begingroup$ I think you're approaching this well. I obtain $$P=\int_{-\infty}^\infty\left(1-\prod_{i=2}^M \Phi(\sigma_i t + \mu_i)\right)\phi\left(\frac{t-\mu_1}{\sigma_1}\right)\,\mathrm{d}t$$ where $\phi$ is the standard Normal density and $\Phi$ is its CDF. That's as far as one can get unless $\mu_1=\cdots=\mu_M$ and $\sigma_1=\cdots=\sigma_M$. I base this opinion on considering simpler versions of the same question ($M=2,3$) that have appeared on this site. $\endgroup$
    – whuber
    Commented Feb 22, 2018 at 21:41
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    $\begingroup$ I obtain $P=\int_{-\infty}^{\infty} \prod_{i} Q(\frac{t-\mu_{i}}{\sqrt{2}\sigma_{i}}) \frac{1}{\sqrt{2\pi}\sigma_{1}} e^{\frac{(t-\mu_{1})^2}{2\sigma_{1}^2}} dt$ in your equation inside CDF shouldn't be ($\frac{t-\mu_{i}}{\sigma_{i}}$)? and by considering standard normal density, we don't miss a $\sigma_{1}$ in denominator of $1/\sqrt{2\pi}\sigma_{1}$? $\endgroup$
    – Niki
    Commented Feb 23, 2018 at 2:03
  • $\begingroup$ If I consider an approximation, can I write it in this way, and is there any ways to make it simpler? $P = \int_{-\infty} ^{\infty} [\prod_{i=2}^{M} Q(\dfrac{t-\mu_{i}}{\sigma_{i}})] \dfrac{1}{\sqrt{2\pi}\sigma_{1}} e^{-\frac{(t-\mu_{1})^{2}}{2\sigma_{1}^{2}}}~dt \leq \int_{-\infty} ^{\infty} \prod_{i=2}^{M} e^{-\dfrac{(t-\mu_{i})^{2}}{2\sigma_{i}^{2}}}\dfrac{1}{\sqrt{2\pi}\sigma_{1}} e^{-\frac{(t-\mu_{1})^{2}}{2\sigma_{1}^{2}}}~dt= \dfrac{1}{\sqrt{2\pi}\sigma_{1}} \int_{-\infty} ^{\infty} \prod_{i=1}^{M} e^{-\dfrac{(t-\mu_{i})^{2}}{2\sigma_{i}^{2}}} ~dt$ $\endgroup$
    – Niki
    Commented Feb 23, 2018 at 3:50

1 Answer 1

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Since $$P=\mathbb{P}(X_1\ge X_2,\ldots,X_1\ge X_n)=\mathbb{P}(X_1-X_2\ge 0,\ldots,X_1-X_n\ge 0)$$ this probability $P$ is the value of a generic $(n-1)$ Normal cdf at $(0,\ldots,0)$, for which there is no closed-form solution. Actually, even for $n=2$ there is no closed-form expression. On the other hand, there exist packages and software that solve this probability computation.

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