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I am having some trouble understanding the steps in the following integral, where $x$ and $y$ are uniform on $[0,1]$.

$$ \int_0^1 \int_0^1 y \, P(y\leq x) \, \mathrm{d}y \, \mathrm{d}x =\int_0^1 x \int_0^xy \, \mathrm dy \, \mathrm dx $$ By using the uniform distribution, I get:

$$\int_0^1 \int_0^1 y \, P(y\leq x) \, \mathrm dy \, \mathrm dx = \int_0^1 \int_0^1 y \, x \hspace{1mm} \, \mathrm dy \, \mathrm dx$$

But why can't I write:

$$= \int_0^1 x \, \mathrm dx \int_0^1 y \, \mathrm dy $$

Can someone help me understand the step here?

Thank you.

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  • $\begingroup$ What are you even trying to find? $\mathsf P(y<x)$ makes little sense, since $y$ and $x$ are not random variables, but the integration terms. Do you mean $\mathbf 1_{y<x}$ , an indicator function -- which equals one when the indicated condition is true, and zero otherwise-- ? $\endgroup$ – Graham Kemp Mar 21 at 23:04
  • $\begingroup$ I'm sorry, it was not clear. $y$ is a random variable. Actually, what I am trying to find is the expected value: $\mathbb{E}(y \hspace{1mm} P(y \leq x))$. $\endgroup$ – Cola Mar 21 at 23:25
  • $\begingroup$ $y$ is either a random variable, xor a term of integration. You cannot use it for both, hence the confusion. $\endgroup$ – Graham Kemp Mar 21 at 23:30
  • $\begingroup$ Thank you. Yes, your answer below is correct. $\endgroup$ – Cola Mar 21 at 23:42
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What are you even trying to find?   $\mathsf P(y<x)$ makes little sense, since $y$ and $x$ are not random variables, but the integration terms.   Do you mean $\mathbf 1_{y<x}$ , an indicator function --which equals one when the indicated condition is true, and zero otherwise-- ?

If so, then you cannot separate the integrals because you must integrate over the domain where the values for $X$ are less than those for $Y$.   There is a clear entanglement.

Thus I suspect that the actual solution is:

$$\begin{align}\mathsf E(Y\mid Y<X) &= \iint_{\Bbb R^2} y~f_{X,Y}(x,y)~\mathbf 1_{y<x}~\mathsf d(x,y)\\[1ex]&= \iint_{\Bbb R^2} \mathbf 1_{0\leqslant y<x\leqslant 1}~y~\mathsf d(x,y)\\[1ex] &= \int_0^1\int_0^x y~\mathrm d y~\mathrm d x \\[1ex] &= \int_0^1 x^2/2~\mathrm d x\\[1ex] &= \tfrac 16 \end{align}$$


I'm sorry, it was not clear. y is a random variable. Actually, what I am trying to find is the expected value: $\Bbb E(y\mathsf P(y≤x))$.

They are either random variables, xor a terms of integration. You cannot use the symbols for both, hence the confusion.

Then as they are random variables, $\mathsf P(y<x)$ is therefore a constant.

$$\begin{align} \mathbb E(y\mathsf P(y\leq x)) &= \mathsf P(y\leq x)\cdot\mathbb E(y)\\[1ex]&=\iint_{\Bbb R^2}\mathbf 1_{t<s}\mathrm d (s,t)~\cdot~\iint_{\Bbb R^2} t~\mathrm d (s,t)\\&=\tfrac 12\cdot\tfrac 12\\&=\tfrac 14\end{align}$$

Are you sure that is what you wanted?

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In the context of the integral, $x$ and $y$ are just numbers, so $P(x<y)$ is simply either 0 or 1 depending on the values of $x$ and $y$.

This is different from $P(X<y)$ where X is a uniform random variable on $[0,1]$. That is equal to $y$.

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