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?

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    $\begingroup$ 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
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    $\begingroup$ 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



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

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

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    $\begingroup$ 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
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    $\begingroup$ 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

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