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Consider some random variables $X_1$, . . ., $X_n$ that are independently and identically distributed with CDF $F(x)$, and some random variables $Y_1, . . ., Y_n$ which are independently and identically distributed with CDF $G(y)$. Suppose further that every $X_i$ is independent of every $Y_i$.

These pairs of random variables yield random sums, $X_1 + Y_1$, . . . , $X_n + Y_n$. I would like to find the probability that one sum is the maximum given that its $X$ value falls below a certain cutoff, i.e.

$$ P(X_1 + Y_1\geq X_j+Y_j \hspace{0.1cm}\forall i\hspace{0.1cm}|\hspace{0.1cm}Y_1 \leq c)$$

Presumably, this is a depends on $F(x)$, $G(x)$, $n$ and $c$.

It is not hard to calculate the probability that $X_1 + Y_1$ is the maximum (unsurprisingly, this equals $1/n$). However, I have been struggling with the case with the constraint and would be very grateful for any help, tips or pointers.

Thanks in advance!

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  • $\begingroup$ From where did all the $U$ come ? $\endgroup$ – Graham Kemp Nov 26 '18 at 23:29
  • $\begingroup$ Apologies, a typo! Now fixed. $\endgroup$ – afreelunch Nov 27 '18 at 0:14
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Yes, it should clearly be that $\mathsf P(X_1+Y_1={\max \{X_i+Y_i\}}_{i=1}^n)=\tfrac 1n$ by symmetry, for continuous random variables, and thus that .

$$\small\begin{split}\mathsf P(X_1{+}Y_1{=}\max_i{\{X_i{+}Y_i\}}\mid X_1{+}Y_1{\leq} u)~\mathsf P(X_1{+}Y_1{\leq}u)&=\tfrac 1n-\mathsf P(X_1{+}Y_1{=}\max_i{\{X_i{+}Y_i\}},X_1{+}Y_1{>}u)\end{split}$$

However, that does not guarantee that $\mathsf P(X_1{+}Y_1{=}\max_i{\{X_i{+}Y_i\}}\mid X_1{+}Y_1{\leq} u)\leqslant \tfrac 1n$.

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  • $\begingroup$ Thanks for the answer and apologies again for the typo. First, to clarify, we are conditioning on $Y_i < c$ not $X_i+Y_i < c$. Can you explain how $P(X_1+Y_1 \geq X_j+Y_j \hspace{0.1cm}\forall i\hspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example? $\endgroup$ – afreelunch Nov 27 '18 at 0:24
  • $\begingroup$ It is the same as above, just with a different partition. But that is more likely to be less than 1/n. I would have to think about it, ... $\endgroup$ – Graham Kemp Nov 27 '18 at 1:45

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