probability of one random variable being greater than another

Using the normal distribution. Let $X \sim N(1, 2)$ and $Y \sim N(2, 3)$ where $N(\mu, \sigma^2)$ denotes the normal distribution with mean $\mu$ and variance $\sigma^2$. $X$ and $Y$ are independent.

What is $P(X>Y)$?

I know that $P(X>Y)$ can be translated to mean $P(X-Y>0)$ and I want to make $X-Y$ into one variable such as $D$. So $P(D>0)$ but how do I subtract the distributions? I tried to do $1-2=-1$ for the mean and then $2-3=-1$ for the variance. I do not understand how this can be because we cannot take the square root of -1 to get the standard deviation.

• What has U got to do with this problem? And is this a homework problem (it is useful to know so that the answer can ensure it focuses on teaching you how to solve the problem) Feb 21, 2013 at 8:21
• Sorry there was more parts to the problem, but this was just the part that I had difficulty understanding. Yes, it is a homework problem. Thanks for the help! Feb 21, 2013 at 21:40
• It's a bad sign when extraneous material appears in questions: it suggests you are merely copying and not thinking, so your readers accordingly spend equally little time thinking about the problem. I have therefore edited out the reference to $U$.
– whuber
Feb 21, 2013 at 22:19
• Sorry I am new to this! But I tried to work out some of the problem in the comment below. Feb 21, 2013 at 22:23

This question is now old enough that I can give you a solution without ruining your homework. As you have pointed out in your question, to compute this probability, you need to find the distribution of $$D=X-Y$$. One of the properties of the normal distribution is that any linear combination of independent normal random variables is also a normal random variable, so this establishes that $$D \sim \mathcal{N}$$, and it remains only to find the mean and variance of this random variable. You have correctly derived the mean, but incorrectly derived the variance.

To obtain the mean and variance of $$D$$ we apply standard rules for the mean and variance of linear functions of random variables. Since the mean is a linear operator and the variance is a quadratic operator, we have:

$$\mathbb{E}(D) = \mathbb{E}(X-Y) = \mathbb{E}(X) - \mathbb{E}(Y) = 1-2 = -1.$$

$$\mathbb{V}(D) = \mathbb{V}(X-Y) = \mathbb{V}(X) + (-1)^2 \mathbb{V}(Y) = 2+3 = 5.$$

Thus, you have the probability:

\begin{aligned} \mathbb{P}(X>Y) = \mathbb{P}(X-Y>0) &= \mathbb{P}(D>0) \\[6pt] &= \mathbb{P} \bigg( \frac{D+1}{\sqrt{5}} > \frac{0+1}{\sqrt{5}} \bigg) \\[6pt] &= 1-\Phi \bigg( \frac{0+1}{\sqrt{5}} \bigg) \\[6pt] &= 0.3273604, \\[6pt] \end{aligned}

where $$\Phi$$ denotes the cumulative distribution function for the standard normal distribution.

• Can I just say more simply that 𝕍(𝑋−𝑌) = 𝕍(𝑋+𝑌) ? Apr 22, 2021 at 22:35
• That is true for uncorrelated random variables, but it is not true in general, so be careful.
– Ben
Apr 22, 2021 at 23:36
• @Ben Can you please explain the logic/basics behind the steps following P(D>0), where you subtract the mean and divide with variance for both RHS, LHS and then reason behind simplifying it further with 1- ....... I assume its because you convert RHS to standard normal variable and then to find P(standard normal variable)> some value, you do 1- "cumulative distribution until the value", isn't it ?
– tjt
Jan 30, 2022 at 4:15
• @tjt: Yes, that is correct.
– Ben
Feb 14 at 20:15

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $$P(X>Y)$$ why not think about $$P(X-Y>0)$$. This is clearly the same probability yes? So now you just need to work out the distribution of $$Z=X-Y$$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $$Y \sim N(1,2)$$ then what is the distribution of $$2Y$$? Well, we double the mean and multiply the variance by $$2^2$$, so $$Y \sim N(2,8)$$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $$Z = X + (-Y)$$ instead?

(In fact this is basically the same argument as pointed out in an older question by Dilip Sarwate: https://stats.stackexchange.com/a/31328/6633)

• Oh sorry, I forgot to mention the work I did. Thanks for the update! Sorry, this is new to me. I know that P(X>Y) can be translated to mean P(X-Y>0) and I want to make X-Y into one variable such as D. So P(D>0) but how do I subtract the distributions? I tried to do 1-2=-1 for the mean and then 2-3=-1 for the variance? I do not understand how this can be because we cannot take the square root of -1 to get the standard deviation. Thanks! Feb 21, 2013 at 22:23
• I love this comment! Please consider moving it into your question as an edit, because it will show people exactly where you're stuck and what you need to know to move forward. (+1 for the question now.)
– whuber
Feb 21, 2013 at 22:26
• @dataznkid1 See this answer to understand why the variances add instead of subtracting. Feb 21, 2013 at 22:26
• Oh, so you add the variances instead. However, is the mean still equal to -1? I am not sure about the negative mean. Do I add this as well? And the variance is equal to 5? So D~N(-1,5)? Feb 21, 2013 at 22:31
• Probably true, @Macro, but if he ever comes back to ask another Q, he'll see that comment. Maybe it'll help, maybe it won't... Sep 5, 2013 at 16:36

$D=X-Y$ is normal with mean $-1$ and variance $2+3$. Knowing the distribution of $D$, you can calculate required probability.

I think this should work.

In general, suppose $$X$$ has distribution function $$\mathbb{G}(x)$$, and $$Y$$ has distribution function $$\mathbb{H}(x)$$ and $$X$$ and $$Y$$ are independent. We need to find the probability $$P(X>Y)$$.

Now, for some constant $$\alpha$$, $$P(Y<\alpha)=\mathbb{H}(\alpha)$$. Therefore, $$P(Y.