Can identically distributed random variables $X$ and $Y$ have $P(X < Y) \geq p$?

Question

Question comes from Joe Blitzstein's "Introduction to Probability".

Let $X$ denote days of the week, encoded as $1, 2, ..., 7$ with equal probabilities. Set $Y = (X + 1)$ mod $7.$ It is easy to see that $Y$ and $X$ are identically distributed. Moreover, $$P(X < Y) = 6/7$$

In general, let $X$ be a random variable with support $\{1, 2, 3, ..., N\},$ and let $Y = (X + 1)$ mod $N$. Similarly to the argument before, $$P(X < Y) = (N-1)/N$$

The problem goes on to ask if it is possible to have $P(X < Y) = 1.$ My argument is that it is possible by $$\lim_{N \to \infty} (N-1)/N = 1$$

Question 1: $X$ is uniformly distributed with each value having probability $1/N.$ Letting $N$ go to inifinity makes individual probabilities $0$, so I am not sure if I can make the argument above.

Question 2: What if $X$ and $Y$ are i.i.d?

I think, if they are i.i.d. we can't make any statement of the form $P(X < Y) \geq p,$ since information about $X$ gives us information about $Y$. For example, letting $p=0.9$ and observing $X=3$ would mean that $P(Y \geq 3) \geq 0.9$. On the other hand, observing $X = 1$ would mean that $P(Y \geq 1) \geq 0.9$, assigning less weight to the $Y \geq 3$ area. Can someone hint at a more rigorous proof?

Answer 1

Partial answer: if $X$ and $Y$ have the same distribution we cannot have $P\{X<Y\}=1$. Proof: this is easy if the common distribution has finite mean: $X<Y$ with probability $1$ and the $E(Y-X)=EY-EX=0$ which is a contradiction since a strictly positive random variable cannot have mean $0$. To handle the general case observe that $\tan ^{-1} Y-\tan ^{-1} X$ is a positive random variable with mean $0$.

If $X$ and $Y$ are i.i.d. with a continuous distribution then $P\{X<Y\}=\frac 1 2$.

Answer 2

For Question 1, note that if $P(X<Y)=1$ and $X$, $Y$ take values in $\{1,\dots,n\}$ then $P(X=n)=0$.

For Question 2, if $X$, $Y$ are iid then $(X,Y)$ has the same distribution as $(Y,X)$ since there is only one joint distribution of $X$, $Y$ for which $X$ and $Y$ are independent. So $P(X<Y)=P(Y<X)$ and so $1=P(X=Y)+2P(X<Y)$, meaning $P(X<Y)\le 1/2$.