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












0














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.



$$ P(X_1 + Y_1geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|hspace{0.1cm}Y_1 leq c)$$



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!






probability probability-distributions







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  • From where did all the $U$ come ?
    – Graham Kemp
    Nov 26 at 23:29










  • Apologies, a typo! Now fixed.
    – afreelunch
    Nov 27 at 0:14
















0














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.



$$ P(X_1 + Y_1geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|hspace{0.1cm}Y_1 leq c)$$



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!






probability probability-distributions







share|cite|improve this question
























  • From where did all the $U$ come ?
    – Graham Kemp
    Nov 26 at 23:29










  • Apologies, a typo! Now fixed.
    – afreelunch
    Nov 27 at 0:14














0












0








0







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.



$$ P(X_1 + Y_1geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|hspace{0.1cm}Y_1 leq c)$$



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!






probability probability-distributions







share|cite|improve this question















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.



$$ P(X_1 + Y_1geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|hspace{0.1cm}Y_1 leq c)$$



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!






probability probability-distributions




probability probability-distributions






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share|cite|improve this question













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edited Dec 2 at 21:03

























asked Nov 26 at 23:22









afreelunch

213




213












  • From where did all the $U$ come ?
    – Graham Kemp
    Nov 26 at 23:29










  • Apologies, a typo! Now fixed.
    – afreelunch
    Nov 27 at 0:14


















  • From where did all the $U$ come ?
    – Graham Kemp
    Nov 26 at 23:29










  • Apologies, a typo! Now fixed.
    – afreelunch
    Nov 27 at 0:14
















From where did all the $U$ come ?
– Graham Kemp
Nov 26 at 23:29




From where did all the $U$ come ?
– Graham Kemp
Nov 26 at 23:29












Apologies, a typo! Now fixed.
– afreelunch
Nov 27 at 0:14




Apologies, a typo! Now fixed.
– afreelunch
Nov 27 at 0:14










1 Answer
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oldest

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0














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 .



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



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






share|cite|improve this answer





















  • 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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
    – afreelunch
    Nov 27 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 at 1:45











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1 Answer
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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 .



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



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






share|cite|improve this answer





















  • 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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
    – afreelunch
    Nov 27 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 at 1:45
















0














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 .



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



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






share|cite|improve this answer





















  • 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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
    – afreelunch
    Nov 27 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 at 1:45














0












0








0






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 .



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



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






share|cite|improve this 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 .



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



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 23:43









Graham Kemp

84.7k43378




84.7k43378












  • 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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
    – afreelunch
    Nov 27 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 at 1:45


















  • 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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
    – afreelunch
    Nov 27 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 at 1:45
















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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
– afreelunch
Nov 27 at 0:24






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 $P(X_1+Y_1 geq X_j+Y_j hspace{0.1cm}forall ihspace{0.1cm}|Y_i < c)$ can exceed $1/n$, perhaps with an example?
– afreelunch
Nov 27 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 at 1:45




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 at 1:45


















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