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Given four random variables, A,B,C,D, chosen independently from the same normal distribution (with mean $\mu$ and standard deviation $\sigma$), I am trying to solve:

$$P[(2+A+B)>(1+B+C) \cap (2+A+B)>(C+D)] $$

The first thing I am doing is combining each of the terms to form a single variable, using summation of normal distributions:

$X_0 = 2+A+B \sim \mathcal N(2+2\mu, 2\sigma^2) $

$X_1 = 1+B+C \sim \mathcal N(1+2\mu, 2\sigma^2) $

$X_2 = C+D \sim \mathcal N(2\mu, 2\sigma^2) $

Now the probability becomes:

$$P(X_0>X_1 \cap X_0>X_2) $$

similar to this question: What is $P(X_1>X_2 , X_1>X_3,... , X_1>X_n)$?

However, the solution to that problem assumed that each $X$ is chosen independently, which may not apply here. I believe that a dependency would exist between $X_1$ and $X_2$, since they share the same selection C. (the same could be said for $X_0$ and $X_1$, but the variables could cancel out in the comparison so it may not matter. $X_1$ and $X_2$, though, are not compared directly). How do I proceed with this to handle the dependencies?

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Note that $$\left(\begin{matrix}X_0\\ X_1\\ X_2\\\end{matrix}\right)=\left(\begin{matrix}2\\1\\0\\ \end{matrix}\right)+\overbrace{\left(\begin{matrix}1 &1 &0 &0\\ 0 &1 &1 &0\\ 0 &0 &1 &1 \\\end{matrix}\right)}^{\mathbf T}\left(\begin{matrix}A\\B\\C\\D\\\end{matrix}\right)$$ which as a linear transform of a Normal vector is again a Normal vector $$\left(\begin{matrix}X_0\\ X_1\\ X_2\\\end{matrix}\right)\sim \mathcal N_3\Bigg(\left(\begin{matrix}2\\1\\0\\ \end{matrix}\right)+\underbrace{\mathbf T\left(\begin{matrix}\mu\\\mu\\\mu\\\mu\\ \end{matrix}\right)}_{\left(\begin{matrix}2\mu &2\mu &2\mu\\\end{matrix}\right)^\text{T}},\sigma^2 \underbrace{\mathbf T\mathbf T^\text{T}}_{\left(\begin{matrix} 2 &1 &0\\ 1 &2 &1\\ 0 &1 &2 \\\end{matrix}\right)} \Bigg)$$ and that again the transform is Gaussian $$Z=\left(\begin{matrix}X_0-X_1\\ X_0-X_2\\\end{matrix}\right)=\overbrace{\left(\begin{matrix}1 &-1 &0\\1 &0 &-1\\ \end{matrix}\right)}^{\mathbf D}\left(\begin{matrix}X_0\\ X_1\\ X_2\\\end{matrix}\right)$$ Hence $$Z\sim \mathcal N_2\Bigg(\underbrace{\mathbf D \left(\begin{matrix}2+2\mu\\1+2\mu\\0+2\mu\\ \end{matrix}\right)}_{\left(\begin{matrix}1&2\\\end{matrix}\right)^\text{T}},\sigma^2 \underbrace{\mathbf D \mathbf T \mathbf T^\text{T}\mathbf D^\text{T}}_{\left(\begin{matrix}2 &2\\2 &4\\\end{matrix}\right)} \Bigg)$$ This brings the question down to finding the distribution of the minimum of two correlated Normal rvs, $\min(Z_1,Z_2)$, since the probability of interest is $$\mathbb P(\min(Z_1,Z_2)>0)$$ and this question is solved here on X validated. With a slight modification since the means and variances are not the same.

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  • $\begingroup$ Thank you for the response. In the second equation down, I see why the means form a 1D vector for the Normal vector, but I don't see why the variances become a 2D matrix. How come the variances are not in a 3x1 vector also? $\endgroup$
    – redantu
    Jul 28, 2020 at 0:39
  • $\begingroup$ @redantu: sorry, I do not understand the question: the mean of a $d$ dimensional vector is a $d$ dimensional vector and the variance of a $d$ dimensional vector is a $d\times d$ matrix. If you are unfamiliar with random vectors, the question should prove delicate to solve. $\endgroup$
    – Xi'an
    Jul 28, 2020 at 6:54
  • $\begingroup$ It looks like I will have to dig into random vectors. I am not familiar with the subject, but I am interested in solving the problem. $\endgroup$
    – redantu
    Jul 28, 2020 at 10:57

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