Find $P(Y_1<Y_3)$?

Question

I am trying to answer this question

Let $Y_1,Y_2,Y_3$ be random variables with pdf $f(x)=2x$ for $0<x<1$. Find $P(Y_1<Y_3)$?

My solution:

Since

$P(Y_1<Y_3)+P(Y_3<Y_1)+P(Y_1=Y_3)=1$

and

$P(Y_3=Y_1)=0$

$P(Y_1<Y_3)=P(Y_3<Y_1)$

then

$P(Y_1<Y_3)=1/2$

Is this correct?

Note: This question was not in English originally. If translated exactly it says "$Y1,Y2,Y3$ are ordered/arranged values ​​for a random sample of size 3". Does this change the question?

Answer 1

Yes indeed, your reasoning is elegant and correct -- for independent random variables.

To confirm, by integration: $$\begin{align}\mathsf P(Y_1<Y_3)&=\int_0^1 \int_0^{y_3} 4y_1y_3~\mathrm d y_1~\mathrm dy_3\\[1ex]&=\int_0^1 2y_3^3~\mathrm dy_3 \\[1ex]&=\tfrac 12\end{align}$$

Answer 2

If translated exactly it says "Y1,Y2,Y3 are ordered/arranged values ​​for a random sample of size 3". Does this change the question?

... a little bit :(. They are Order Statistics, and must be indicated with $Y_{(i)}$

As already noted in a previous comment, $\mathbb{P}[Y_{(1)}<Y_{(3)}]=1$ as $Y_{(1)}<Y_{(3)}$ a.s.

To confirm the result by integration, let's find the joint density

$f_{Y_{(1)}Y_{(3)}}(u,v)=24uv(v^2-u^2)$

$u,v \in [0;1]$

and then proceed by integration

$$\mathbb{P}[Y_{(1)}<Y_{(3)}]=\int_0^1 \int_0^v [24uv^3-24u^3v]dudv =...=1$$