# Conditional Probability with Normal Distributions?

Let's say that I have $3$ independent random normal variables, $A$, $B$ and $C$.

They all have a standard deviation of $17.526$, while $A$ has a mean of $143$, $B$ of $139$, and $C$ of $129$.

My question is, how would I go about calculating $P(A>C\mid A>B)$? I know how to calculate $P(A>C)$ and $P(A>B)$, but it's the conditional part I'm struggling with. I'm not sure how I would manage to apply Bayes' theorem to this, so I'm stuck.

If it helps speed-wise, I have $P(A>C)$ to be $0.713912$, and $P(A>B)$ to be $0.564105$ with the difference distribution in each case having a stdev of $24.785$.

Thanks.

-
Duplicate of this question on math.SE. – Dilip Sarwate Aug 29 '14 at 18:08

Given $3$ independent normal random variables $A, B$, and $C$, we have that $$P\{A > C \mid A > B\} = \frac{P\left(\{A > C\}\cap \{A > B\}\right)}{P\{A > B\}} \tag{1}$$ where the denominator of the fraction on the right side of $(1)$ is easy to compute by observing that $B-A \sim N(\mu_B-\mu_A, \sigma_B^2+\sigma_A^2)$, and so $$P\{A > B\} = P\{B-A < 0\} = \Phi\left(\frac{\mu_A-\mu_B}{\sqrt{\sigma_B^2+\sigma_A^2}}\right).\tag{2}$$ The OP has made this observation and done this computation already. However, the numerator of the fraction on the right side of $(1)$ is not easily computed, and numerical integration might be necessary. One approach using conditional distributions is as as follows. Given that $A=a$ \begin{align} P\left(\{A > C\}\cap \{A > B\}\mid \{A = a\}\right) &= P\left(\{a > C\}\cap \{a > B\}\right)\\ &= P\{a > C\}P\{a > B\} & \scriptstyle{B~ \text{and}~C~\text{are independent}}\\ &= \Phi\left(\frac{a-\mu_C}{\sigma_C}\right)\Phi\left(\frac{a-\mu_B}{\sigma_B}\right). \end{align} We can now use the law of total probability to write that \begin{align} P\left(\{A > C\}\cap \{A > B\}\right) &= \int_{-\infty}^{\infty} P\left(\{A > C\}\cap \{A > B\}\mid \{A = a\}\right)f_A(a)\,\mathrm da\\ &= \int_{-\infty}^{\infty} \Phi\left(\frac{a-\mu_C}{\sigma_C}\right)\Phi\left(\frac{a-\mu_B}{\sigma_B}\right) \frac{1}{\sigma_A}\phi\left(\frac{a-\mu_A}{\sigma_A}\right)\,\mathrm da. \end{align} To the best of my knowledge, there is no "closed-form" expression for this probability and the value of the above integral has to be computed via numerical integration. Using the value provided by the simulation by @MasatoNakazawa with the value $P(A>B) = 0.564\ldots$ provided by the OP, it would appear that for the values of the parameters in the OP's problem, $$P\left(\{A > C\}\cap \{A > B\}\right) \approx 0.468\ldots$$

Crude bounds on the above probability can be obtained for those not inclined to simulate or numerically integrate. We have $$P\left(\{A > C\}\cap \{A > B\}\right) \leq \min \left\{P\{A > C\}, P\{A > B\}\right\} = 0.564105$$ Also, the complementary probability $P\left(\{A < C\}\cup \{A < B\}\right)$ is bounded above as $$P\left(\{A < C\}\cup \{A < B\}\right) \leq P\{A < C\} + P\{A < B\} = 0.721983$$ and so we have that $$0.278017 \leq P\left(\{A > C\}\cap \{A > B\}\right) \leq 0.564105.$$

-
+1. Observing that the answer is invariant under changes of scale and location allows a simpler expression for the formulas because you may assume $\sigma_A=\sigma_B=\sigma_C$ (as stated in the problem) $=1$ and (say) $\mu_A=0$. That makes the solution a function of the two variables $\mu_B=139/17.526$ and $\mu_C=129/17.526$. – whuber Aug 31 '14 at 19:23
@whuber Thanks for the upvote. I agree that in this particular case of equal variances, the result could have been expressed more simply. However, I chose to give a somewhat more general answer (to dot the i's and cross the t's, so to speak) so that the answer could be used directly in situations when the variances were not equal. – Dilip Sarwate Aug 31 '14 at 19:50

I just want to supplement @Comp_Warrior's great answer (an empirical estimate of $P(A > max(B, C))/P(A>B)$):

s <- 17.526
N <- 10000

set.seed(1)
A <- rnorm(N, 143, s)
B <- rnorm(N, 139, s)
C <- rnorm(N, 129, s)

sum(A > pmax(B, C))/sum(A>B) ## ~0.83

-

You can use the definition of conditional probability:

$$P(X|Y)=\frac{P(X,Y)}{P(Y)}$$

$$P(A>C|A>B) = \frac{P(A>C,A>B)}{P(A>B)} = \frac{P(A>\max(B,C) )}{P(A>B)}$$
You could find the distribution of $\max(B,C)$ and then find the probability of the numerator.
I have deleted my previous comments. The last line of your answer is the hard part. To the best of my knowledge, simulation or numerical integration is needed to compute $P(A > \max(B,C))$. – Dilip Sarwate Aug 30 '14 at 12:49