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I have physically measured two random variables X and Y and determined their respective mean and standard deviation. Both variables have a gaussian distribution. Now I wish to calculate the probability that a given value of X will be less than a given value of Y.

For example, if variable X has a mean of 100 and a standard deviation of 10 and variable Y has a mean of 120 and a standard deviation of 15, what is the probability of X being less than Y given Y=120?

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Are you asking for the probability that X<Y, or for the probability that X<Y conditionally on Y=y, for some given value y? These do not coincide. – Did May 5 '12 at 11:11
I'm not sure....however it may help if I provide more context. The specific situation concerns swimming race times. One competitor (competitor A) has a mean of 100seconds with a SD of 10s and another competitor (competitor B) has a mean of 120seconds with a SD of 15s. I'm assuming that the swim times are independent (even if this is not strictly valid in reality). So, in any given race, what is the probability of competitor A finishing before competitor B? – Jason May 5 '12 at 11:27
Then your question is to determine the probability that X<Y, in contradiction with the second paragraph of your post. – Did May 5 '12 at 12:21
up vote 1 down vote accepted

This problem cannot be answered unless we have additional information. But it can be answered if we assume that $X$ and $Y$ are independent. You will have to judge whether the assumption of independence is reasonable in your situation.

If $X$ and $Y$ are independent, then the difference $W=X-Y$ is Gaussian, mean $100-120$, and variance $\sigma^2$, where $\sigma^2=10^2+15^2$.

So (under the assumption of independence of $X$ and $Y$), we know that $W=X-Y$ is Gaussian with mean $-20$ and standard deviation $\sqrt{325}$. Now it is a calculation to find $P(W\lt 0)$.

Remark: Let $X$ and $Y$ be independent Gaussian, with mean $\mu_X$ and $\mu_Y$, and variance $\sigma_X^2$ and $\sigma_Y^2$ respectively. Let $W=aX+bY$ where $a$ and $b$ are constants. Then $W$ is Gaussian, mean $a\mu_X+b\mu_Y$ and variance $a^2\sigma_X^2 +b^2\sigma_Y^2$.

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  • The probability that [X<Y] conditionally on [Y=120] is the probability that [X<120] conditionally on [Y=120].
  • Assume that X and Y are independent. (The question cannot be solved without a hypothesis on the dependence of X and Y.) Then, this is the (unconditional) probability that [X<120].
  • If X is normal with mean 100 and standard deviation 10, then X=100+10Z with Z standard normal, hence [X<120]=[Z<2] and P(X<Y|Y=120)=P(Z<2)=97.7%.
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