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I am trying to find some expression for this integral, ideally as a function of the constants. I can't put the full integral in the title as it's too long.

I doubt a closed form exists, but I'm even stuck with how I would go about it.

$$\frac{1}{4\sqrt{2\pi \sigma_3}}\int_{-\infty}^\infty \exp\left(-\frac{(\mu_3-z)^2}{2\sigma_3}\right)\left(1+\text{Erf}\left[\frac{\mu_2-z}{\sqrt{2}\sqrt{\sigma_2}}\right]\right)\left(1-\text{Erf}\left[\frac{\mu_1-z}{\sqrt{2}\sqrt{\sigma_1}}\right]\right)dz$$

This was derived when considering $P(Y<Z<X)$ where $Y,Z,X$ are random variables that all follow a normal distribution, but with distinct parameters.

blueteethbass
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  • From your integral, tt seems to me that $X,Y,Z$ are independant, doesn't it? – NN2 Feb 19 '21 at 18:32
  • @NN2 Yes, the assumption is that they're all independent so the joint density is just the product of the marginals – blueteethbass Feb 19 '21 at 19:31
  • It is done. I posted the answer on your other question, so maybe you should close this as duplicate? – Sal Feb 19 '21 at 22:27

1 Answers1

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Partial Answer: If we assume independence, then you have $$ U := X-Z \sim N(\mu_X-\mu_Z, \sigma^2_X + \sigma^2_Z) $$ and $$ V := Z-Y \sim N(\mu_Z-\mu_Y, \sigma^2_Z + \sigma^2_Y) $$ and further, $$ \text{Cov}(U, V) = -\text{Cov}(Z,Z) = -\sigma^2_Z $$ so that $$ \begin{bmatrix} U\\V \end{bmatrix} \sim N \left ( \begin{bmatrix} \mu_X-\mu_Z\\\mu_Z-\mu_Y \end{bmatrix}, \begin{bmatrix} \sigma^2_X+\sigma_Z^2 & -\sigma^2_Z\\-\sigma^2_Z & \sigma^2_Y+\sigma_Z^2 \end{bmatrix} \right ) $$ We therefore have $$ P(Y<Z<X) = P(U > 0, V>0). $$ If $U,V$ were correlated standard Gaussians with correlation $\rho$, then we would have $$ P(U>0, V>0) = \frac{1}{4} + \frac{1}{2\pi} \arcsin \rho $$ based on the result in this question. So perhaps you can use the result there to work out something nicer for your problem?

WeakLearner
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