Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For three independent normally distributed continuous random variables X, Y, and Z (each with its own mean and standard deviation), I need a way to calculate

$P(Y \geq X, Y \leq Z)$

I know that I can do this by the following:

$P(Y \geq X, Y \leq Z) = P(Y \geq X) \cdot P(Y \leq Z | Y \geq X)$

I am able to calculate $P(Y \geq X)$ using the following relation: $P(Y \geq X) = P(Y - X \geq 0)$

However, I'm having trouble calculating precisely $P(Y \leq Z | Y \geq X)$

share|improve this question
Hint: Conditioned on $Y = y$, the event $\{Y \geq X, Y \leq Z\}$ has conditional probability $$P\{Y \geq X, Y \leq Z \mid Y = y\} = P\{X \leq y, Z \geq y \mid Y = y\} = \Phi((y-\mu_X)/\sigma_x)(1 - \Phi((y-\mu_z)/\sigma_Z)).$$ Multiply by the density of $Y$ and integrate. – Dilip Sarwate May 24 '12 at 2:39
Is this a homework problem? Darn it. I have an answer I'm proud of, but I'm not sure if I ought to post it. – Cyan May 24 '12 at 3:33
Just to check - are you calculating $P(x \leq y \leq z)$ or do you already know that $x \leq z$ and are calculating the probability that $y$ is inbetween? – jbowman May 24 '12 at 3:56
Thanks Dilip. I will try that. :) – Abey May 24 '12 at 4:03
Cyan, no it's not homework, it's just a problem that was driving me mad. – Abey May 24 '12 at 4:04
up vote 7 down vote accepted

One relatively easy approach is to consider $X$, $Y$, and $Z$ as having a joint multivariate normal distribution.

$\left[\begin{array}{c}X\\Y\\Z\end{array}\right]\sim\mathrm{MVN}\left(\left[\begin{array}{c}\mu_{X}\\\mu_{Y}\\\mu_{Z}\end{array}\right],\left[\begin{array}{ccc}\sigma_{X}^{2} & 0 & 0\\0 & \sigma_{Y}^{2} & 0\\0 & 0 & \sigma_{Z}^{2}\end{array}\right]\right)$


$\left[\begin{array}{c} U\\ V\end{array}\right]=\left[\begin{array}{c} X-Y\\ Z-Y\end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0\\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{c} X\\ Y\\ Z\end{array}\right]$

Then by standard results on affine transformations of multivariate normal distributions,

$\left[\begin{array}{c} U\\ V\end{array}\right]\sim\mathrm{MVN}\left(\left[\begin{array}{c} \mu_{X}-\mu_{Y}\\ \mu_{Z}-\mu_{Y}\end{array}\right],\left[\begin{array}{cc} \sigma_{X}^{2}+\sigma_{Y}^{2} & \sigma_{Y}^{2}\\ \sigma_{Y}^{2} & \sigma_{Z}^{2}+\sigma_{Y}^{2}\end{array}\right]\right)$

And since $P(Y \geq X, Y \leq Z) = P(U \leq 0, V \geq 0)$, you want the probability mass of this bivariate distribution in the second quadrant. This is not analytically solvable in general, but is easy to compute. If $\mu_X = \mu_Y = \mu_Z$, then there is an analytical expression (from equation 73 here):

$P(U \leq 0, V \geq 0) = \frac{1}{2} \cos^{-1}\left(\frac{\sigma^2_{Y}}{\sqrt{(\sigma^2_{X} + \sigma^2_{Y}) (\sigma^2_{Z} + \sigma^2_{Y})}}\right)$.

Added: Here's R code to compute the probability.

mu_x <- -1.4
mu_y <- 2
mu_z <- 1.7
mu_vec <- c(mu_x- mu_y, mu_z - mu_y) 
var_x <- 9
var_y <- 9
var_z <- 16
Sigma <- var_y + matrix(c(var_x, 0, 0 , var_z), nrow = 2)
pmvnorm(lower = c(-Inf, 0), upper = c(0, Inf), mean = mu_vec, sigma = Sigma)
share|improve this answer
Thank you Cyan. Although I'm not experienced on the level you post about, I will study upon it. Much appreciated. :) – Abey May 24 '12 at 4:20
What pieces are you missing? Maybe I can point you to some resources. – Cyan May 24 '12 at 4:26
Honestly, I am not very familiar with the standardized ways of expressing these concepts. Thus, the meaning of those first three parts is unclear to me. I understand the last line; however, the means of the three distributions are not necessarily the same. – Abey May 24 '12 at 4:34
The bivariate normal distribution might be a good starting point -- the linked page has nice animations. Then you'll need to know about vectors (especially in Cartesian space) and matrices. (cont'd) – Cyan May 24 '12 at 5:11
After that you'll be ready to tackle random vectors and the multivariate normal distribution. – Cyan May 24 '12 at 5:11

I might just make many draws from the distribution and calculate the rate that the event you are interested in occurs. In R:

 sum(x<y & y >z )/N

It is just an estimation so maybe do it a couple times. Quick and dirty

share|improve this answer
The final line of code should be "sum(x<y & y<z )/N". – Cyan May 31 '12 at 13:04
This is a great way to check an answer. But did you notice the word "precisely" in the question? – whuber May 31 '12 at 14:13
Thanks Seth, and ditto on good idea to check the answer. – Abey May 31 '12 at 15:11
I noticed the word 'precisely', I just ignored it. Just kidding. Both cyan's and this method can be computed with any level of precision. – Seth May 31 '12 at 15:15
Up to a point Seth: the precision is inherently limited by your computing capabilities and lifetime of the universe. Remember, the SD of a simulated estimate scales only as $N^{-1/2}$. That will make it prohibitively difficult to attain more than about six significant figures; your example only gets about four sig figs. (Every once in a while even six sig figs is not good enough as a check, so this is not entirely nit-picking.) – whuber May 31 '12 at 18:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.