# A question about conditional probability

Assume $X$ is discrete random variable. It has some distribution on integers $0,1,...,m$

Then is $P(X>k | X=m) \ge P(X>k)$ true? If it is, how to prove it in a rigorous way?

For me, it is tempting to think that it is true because given $X$ taking the maximum value, the probability of $X>k$ should be higher. But I can't figure out how to prove my thought.

If $m$ is the maximum possible value of $X$, then the conditional probability in your problem can only take values zero and one

$$P(X > k \mid X = m) = 1 \text{ if } k < m$$ $$P(X > k \mid X = m) = 0 \text{ if } k \geq m$$

In the first case, the inequality is obviously true because all probabilities are bounded above by one. In the second case

$$P(X \geq k) = 0$$

since $k \geq m$, and $m$ is the maximum possible value of $X$.

• This is the same as my answer which I was writing while you were answering. Oct 17, 2017 at 5:29

For any $k < m$ the $P(X>k| X=m)=1$ which is certainly greater than $P(X>k)$ (assuming each integer from 0 to m has positive probability). If $k=m$ $P(X>k)=0$ and so does $P(X>k|X=m)$.

• $P(X>k \mid X=m) = 1 < P(X>k)$. So, $P(X>k) > 1$? Oct 17, 2017 at 11:12
• No I said P(X>k|X=m)=1 >P(X>k)=0 when k=m. In the first case I meant to write P(X>k|X=m) =1 > P(X>k) when k<m). I wrote "less" when of course I meant "greater". i have corrected it. Oct 17, 2017 at 14:12