# What is probability that one normal random variable is max of three normal random variables?

I have three independent random variables, $X_1$, $X_2$, and $X_3$, and each random variable has different mean. (And I assume they have the same variance). I would like to know how I can get the probability that $X_1$ is the max of all three. And help would be appreciated.

• Do you know the distributions? Commented May 16, 2016 at 18:23
• I assume they all have the normal distributions with same variance but different means. Commented May 16, 2016 at 18:29
• Do values of X1 correspond to values of X2 & X3? Commented May 16, 2016 at 18:30
• gung, I don't understand your question. Could you clarify? Commented May 16, 2016 at 18:39
• Closely related: stats.stackexchange.com/questions/87241 (generic formulation), stats.stackexchange.com/questions/173064 (two independent normals), stats.stackexchange.com/questions/139072 (two correlated normals), etc. Note that $X_2-X_1$ and $X_3-X_1$ are correlated Normals and $X_1$ is the max of all $X_i$ if and only if the larger of these two differences is non-positive. Thus, you may directly apply the answer in the last link.
– whuber
Commented May 16, 2016 at 20:04

Assume their means are $\mu_1,\mu_2,\mu_3$, with the same standard deviation, $\sigma$. Then the probability that the first random variable is the max is:
$$P(X_1>X_2,X_1>X_3)=\int_{-\infty}^\infty P(x>X_2|X_1=x)P(x>X_3|X_1=x)f(x)dx$$
where $f(x)$ is the density of $X_1$. It's doubtful whether you can simply this further.
• Since the variables are independent this simplifies to $\int_{-\infty}^{\infty} {P(X_2 < x) P(X_3 < x) f(x) dx}$ Commented Apr 24, 2023 at 1:43