# Find the probability that X + Y is maximal given that Y < c.

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!

• From where did all the $U$ come ? – Graham Kemp Nov 26 '18 at 23:29
• Apologies, a typo! Now fixed. – afreelunch Nov 27 '18 at 0:14

## 1 Answer

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 $$\mathsf P(X_1{+}Y_1{=}\max_i{\{X_i{+}Y_i\}}\mid X_1{+}Y_1{\leq} u)\leqslant \tfrac 1n$$.

• 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? – afreelunch Nov 27 '18 at 0:24
• 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, ... – Graham Kemp Nov 27 '18 at 1:45