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I'm looking for some SAS code that will compute the probability of one value (randomly taken from a normal distribution) being greater than another value (randomly taken from a another normal distribution) given the means and standard deviations of both distributions are known. Thank you

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If this is a question of implementation, then perhaps this R code can help you in some way, since I do not have SAS:

n <- 10000000
mean1 <- 1; var1 <- 3
mean2 <- 0; var2 <- 1 # defining parameters

X <- rnorm(n, mean = mean1, sd = sqrt(var1))
Y <- rnorm(n, mean = mean2, sd = sqrt(var2)) #generating random variables

Z <- (X<=Y) #creates logical vector
sum(Z) / length(Z) #I get about 0.308669

Or perhaps you find it easier to use that linear transformations of normal random variables are again normal and use a built in function? For example, if

$$ X \sim N(\mu_1,\sigma_1^2) \text{ and } Y \sim N(\mu_2, \sigma_2^2) \quad \Longrightarrow \quad X \pm Y \sim N(\mu_1 \pm \mu_2, \sigma_1^2 + \sigma_2^2), $$

then $$ P(X \leq Y) = P(X-Y \leq 0) = P\left(\frac{(X-Y)-(\mu_1-\mu_2)}{\sigma_1 + \sigma_2} \leq \frac{-(\mu_1-\mu_2)}{\sigma_1 + \sigma_2} \right) = \Phi \left( \frac{-(\mu_1-\mu_2)}{\sigma_1 + \sigma_2} \right), $$

where $\Phi(\cdot)$ is the cumulative distribution function for the standard normal distribution and we get the following in R (which is nicer):

pnorm(-(mean1-mean2)/sqrt(var1+var2) #I get about 0.308538

I hope this helps with your implementation, please report back.

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  • $\begingroup$ This worked great! thanks! $\endgroup$ Jun 27 '19 at 14:41
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    $\begingroup$ You are welcome, if this answers your question, you should then close it. $\endgroup$
    – K. Brix
    Jun 28 '19 at 8:22

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