2
$\begingroup$

If $X_1,X_2$ are dependent but identically distributed, it seems obvious that $P(X_1+X_2\geq2C) \leq P(X_1\geq C)=P(X_2\geq C)$. At least if we additionally assume that the joint distribution is symmetric, i.e. $P(X_1\leq a, X_2\leq b) =P(X_2\leq a, X_1\leq b)$.

But how would I prove this?

$\endgroup$
3
  • 1
    $\begingroup$ I assume the 'seems obvious' is because it would hold if $X_1$ and $X_2$ were perfectly correlated, and intuitively it seems that perfect correlation should maximise $X_1+X_2$. $\endgroup$ Commented May 16, 2022 at 3:40
  • 1
    $\begingroup$ I'm pretty sure this exact question was asked and answered about a year ago--it's just hard to find with a search. Here is a generalization, though, that includes this question as a special case: stats.stackexchange.com/questions/525638. $\endgroup$
    – whuber
    Commented May 16, 2022 at 12:42
  • $\begingroup$ Yes that was my intuition @ThomasLumley ! Always humbling to learn that it was totally wrong! $\endgroup$
    – Jome
    Commented May 16, 2022 at 18:14

1 Answer 1

3
$\begingroup$

Counterexample

$$P(X_1 = 3, X_2 = 7) = P(X_1 = 7, X_2 = 3) = 0.5$$

Then $$P(X_1 \geq 4) = 0.5$$ and $$P(X_1 + X_2 \geq 8) = 1$$

Thus in this case $P(X_1 \geq c) < P(X_1 + X_2\geq c)$


Below is a little sketch that describes which intuition I used to find these points.

If there are relatively many points in the triangular regions where $X_1 + X_2 > 2c$ while $X_1 < c$ or $X_2 < c$ then you get a counter example to the inequality.

intuitive sketch

$\endgroup$
6
  • 4
    $\begingroup$ The distribution does not need to be negatively correlated or even dependent. For instance, imagine a pair of Bernoulli variables with with $$P(X_1,X_2 = 0,0) = P(X_1,X_2 = 1,0) = P(X_1,X_2 = 0,1) = P(X_1,X_2 = 1,1)$$ and consider $P(X_1 \geq 0.1) = 0.5$ with $P(X_1 + X_2 \geq 0.2) = 0.75$ $\endgroup$ Commented May 16, 2022 at 8:26
  • $\begingroup$ The sketch reminds me an awful lot of stats.stackexchange.com/a/564911/919 ;-). $\endgroup$
    – whuber
    Commented May 16, 2022 at 12:44
  • $\begingroup$ @whuber I must say that this visual great and your answer over there (including the visual) might be better than mine (not including the visual). I actually don't even remember anymore that question and me answering it. I will have to review that other post again if I get the time. $\endgroup$ Commented May 16, 2022 at 13:01
  • 1
    $\begingroup$ It is good to read back what I wrote a while ago. I sometimes don't get it anymore myself so how should the reader get it, and in this case the post is only three months old $\endgroup$ Commented May 16, 2022 at 13:07
  • $\begingroup$ Forgetting stuff you posted is the curse of posting a lot of stuff :-). $\endgroup$
    – whuber
    Commented May 16, 2022 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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