This is a two part question. I apologize if the title and tags are vague. Please edit if a more suitable title or tags are appropriate.

Part 1

Ok, so if $X$ and $Y$ are independent, continuous random variables with range $\mathbb{R}$, then we can write (assuming pdfs exist):

\begin{align*} P(Y \gt X) &= \int_{\{(x,y) \in \mathbb{R}^2 : \; y \gt x\}} p_{X,Y}(x,y) \; dx dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^y p_X(x)p_Y(y) \; dx dy \\ &= \int_{-\infty}^{\infty} P_X(X \leq y) \; p_Y(y) \; dy \\ &= E_Y[P(X \lt Y|Y)] \end{align*}

Is the last step valid, and how would one read this out loud / explain it intuitively?

Part 2

Next, take $Y$ to be Bernoulli with range $\{0, 1\}$ and $X$ with the same range as before. However, $X$ and $Y$ are now not independent. I want to consider the distributions, $X|Y=0$ and $X|Y=1$, and proceed like the previous case with $X|Y=0$ taking the place of $X$ and $X|Y=1$ taking the place of $Y$, but I'm kind of lost in notation and understanding.

  1. How do we write that $X|Y=1$ is independent of $X|Y=0$? What is the notation for the joint distribution?
  2. Does it even make sense that $X|Y=1$ and $X|Y=0$ could be independent? In words, it seems like it's feasible: the distribution of the value of $X$ given that $Y=1$ might not tell you anything about the distribution of the value of $X$ given that $Y=0$.
  3. $P(X|Y=1 \gt X|Y=0)$ seems horribly confusing. What is the notation for this?

Edit: update for Part 2

What I wrote for part was gibberish. So let me update with what my intentions were.

Consider four random variables: $X_1,X_2,Y_1,Y_2$. The ranges of $X_1$ and $X_2$ are both $\mathbb{R}$, while the ranges of $Y_1$ and $Y_2$ are both $\{0,1\}$. Similar to Part 1, I am interested in computing:

$$ P(X_1 > X_2 | Y_1=1,Y_2=0) $$

I was curious what assumptions on the $X_1,X_2,Y_1,Y_2$ allow for a similar calculation. It turns out that we must have both that $(X_1,Y_1)$ is independent of $(X_2,Y_2)$ and $Y_1$ is independent of $Y_2$. Under these assumptions,

$$ P(X_1 > X_2 | Y_1=1,Y_2=0) = \int_{\mathbb{R}} P(X_2 \lt x_1 | Y_2=0) p(x_1|Y_1=1) dx_1 $$

  • 1
    $\begingroup$ In Part 2 (2), there must be a typographical error. In Part 2 (3), yes it's horribly confusing--what do you mean? $\endgroup$
    – whuber
    Aug 11, 2015 at 0:58
  • $\begingroup$ Unfortunately, there is no random variable defined as $X|Y=1$. So the whole Part 2 does not make sense within probability theory. $\endgroup$
    – Xi'an
    Nov 1, 2015 at 14:21
  • $\begingroup$ @Xi'an yes, my part 2 was not well thought-out. I've clarified what I intended in the edit above. $\endgroup$
    – Chester
    Nov 1, 2015 at 15:31

1 Answer 1


Part 1: this is valid, and it would be entirely reasonable to go directly to the last equality, as you are basically rederiving a special case of the law of total expectation. Recall that for random variables $A$ and $B$ we can write $\text{E}(A) = \text{E}[\text{E}(A \mid B)]$. Since probabilities can themselves be viewed as expectations your equation follows from this.

Part 2: you're not being very clear about the underlying probability model, so it's difficult to answer this. In particular, what does it mean to talk about $X \mid (Y = 0)$ and $X \mid (Y = 1)$ being independent? Are we sampling $X$ under the condition that $Y = 1$, and then taking another realization of $X$ under the event $Y = 0$? These random variables could be either dependent or independent, regardless of the joint distribution of $X$ and $Y$. But, I suspect that this isn't what you mean.

  • $\begingroup$ Basically, I want to apply the result of Part 1 to the random variables $X|Y=0$ and $X|Y=1$ in place of $X$ and $Y$ in Part 1 respectively (overloading variable names, sorry). You are correct in your interpretation: I'm interested in the probability that a realization of the value of $X$ given $Y=1$ is larger than a realization of the value of $X$ given that $Y=0$. To this end, I need to consider (in)dependence of these random quantities, which I think coincides with your last couple of sentences. $\endgroup$
    – Chester
    Aug 11, 2015 at 3:24
  • $\begingroup$ I guess I'm just caught up on notation. Usually, we just say two random vars, $A$ and $B$, are independent if $P(A,B) = P(A)P(B)$. In this case, $A:=X|Y=1$, $B:=X|Y=0$. Is there a nice or accepted way of writing that these quantities are independent? $\endgroup$
    – Chester
    Aug 11, 2015 at 3:27
  • $\begingroup$ Hmm. I'm not sure what you mean by that last equality, or where the RHS comes from. Let me try to be more clear: $X$ is a continuous rv. $Y$ is Bernoulli. Define the new rv's, $A = X|(Y=1)$ and $B = X|(Y=0)$. This is a valid thing to do, no? From Part 1, $P(A \gt B) = E[ P(A \gt B |A) ]$. I'm simply asking how this can be written without introducing the new variables, $A$ and $B$? Furthermore, I'm asking how to write the intermediate steps without introducing $A$ and $B$, as well. $\endgroup$
    – Chester
    Aug 11, 2015 at 13:06
  • $\begingroup$ Ok, I misread the post and thought $X$ was also Bernoulli. I'll repost a corrected comment. $\endgroup$
    – dsaxton
    Aug 11, 2015 at 13:19
  • $\begingroup$ If you want to apply the idea from the first part (which by the way should probably use the notation $E[P(X<Y \mid Y)]$; $P_X(X \leq Y)$ doesn't make a lot of sense) then you would write this expectation as $P(Y=1)P(X<Y \mid Y=1)+P(Y=0)P(X<Y \mid Y=0)$. We aren't introducing any new r.v.'s here, everything is in terms of $X$ and $Y$. $\endgroup$
    – dsaxton
    Aug 11, 2015 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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