2
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

4
$\begingroup$

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 $X$.

$\endgroup$
1
  • $\begingroup$ This is the same as my answer which I was writing while you were answering. $\endgroup$ Oct 17, 2017 at 5:29
0
$\begingroup$

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 $P(X>k|X=m)$.

$\endgroup$
2
  • $\begingroup$ $P(X>k \mid X=m) = 1 < P(X>k)$. So, $P(X>k) > 1$? $\endgroup$ Oct 17, 2017 at 11:12
  • $\begingroup$ 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. $\endgroup$ Oct 17, 2017 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.