Suppose I have two discrete random variables $X$ and $Y$ defined on the same sample space. Intuitively I understand that $P(X<Y)$ does not imply that $P(X^{2} < Y^{2})$, but I can't think of any explicit example, however simple, that demonstrates this – can anyone help please?

# Example of a situation when $P(X<Y)$ is not equal to $P(X^{2} < Y^{2})$

probabilityprobability distributionsrandom variables

#### Related Solutions

It seems that one usually use the set of all possible values of X as the sample space of random variable X.

Well, your post is an excellent example of why one should not. Actually, the reason why some introductory probability courses in some countries focus so much on the sample space and (even worse) why they recommend to take as sample space the image set of $X$ is a real mystery since:

- this option is ludicrous in itself,
- it leads to confusion for anybody actually thinking about the problem,
- it is completely abandoned (rightly so) in later, higher, courses (not to mention that probabilists themselves refute it).

Terence Tao's citation in Qiaochu Yuan's answer (and t.b.'s very short comment on his own answer) on the page you link to, all say it well so I will be brief. The take-home message is that, in most situations, *there exists* some probability space $(\Omega,\mathcal F,P)$ such that the family of random variables one is interested in can be defined simultaneously, as long as their joint distribution are compatible. Kolmogorov's consistency theorem gives you that, in a much wider setting than you would actually care for, and this is it. The nature of $(\Omega,\mathcal F,P)$ is left unspecified (although one knows that $[0,1]$ with its Borel sigma-algebra would fit an awful lot of situations) and, more importantly, it is irrelevant. All that counts is that *some* such probability space $(\Omega,\mathcal F,P)$ exists. Once one knows it does, one can turn to the actual probability questions one wants to solve.

Let us say that your random experience is to throw two 6-side fair die. Let $X$ be the value of the first die and $Y$ be the value of the second die.

Consider the event $A =$ the total outcome is greater than $9$

We now want to calculate $P(A)$. Consider the partition of $A$ into $\{A_{10}, A_{11}, A_{12}\}$ where $A_{i}$ is the event where the total outcome is $i$.

What we want now is $P(A)$. But if that happens, it either happened because $A_{10}$ happened, because $A_{11}$ happened or because $A_{12}$ happened. Thus

$$P(A) = P(A_{12}) + P(A_{11}) + P(A_{10})$$

How to calculate each $P(A_i)$? Let us start with $P(A_{10})$.

We know that $X \geq 4$ because the dice can only roll up to 12, since the largest value one can attain is only $6$. Thus $P(A_{10})$ depends on the probabilities $P(Y \geq 4 | X = 6), P(Y \geq 5 | X = 5), P(Y \geq 6 | X = 4)$. In what way though? Well, the law of total probability tells us:

$$P(A_{10}) = P(Y \geq 4 | X = 6)\cdot P(X = 6) + P(Y \geq 5 | X = 5)\cdot P(X = 5) + P(Y \geq 6 | X = 4)\cdot P(X = 4)$$

With similar trains of thought we see that

$$P(A_{11}) = P(Y \geq 5 | X = 6)\cdot P(X = 6) + P(Y \geq 6 | X = 5)\cdot P(X = 5)$$

and

$$P(A_{12}) = P(Y \geq 6 | X = 6)\cdot P(X = 6)$$

Which we can then add up together to find $P(A)$.

Is it now clearer to you?

## Best Answer

Let $X$ be constantly $-1$ and $Y$ be constantly $0$. Then $P(X<Y) = 1 $ and $P(X^2 <Y^2) = 0$