Mixing
probability parameter kf $t$
Suppose we define a probability on some sample space, dependent on a parameter $t$. Let $E$ and $F$ be two events. Suppose that for each of these events, their probability is an increasing function of $t$.
Does this mean that the probability of the union of $E$ and $F$ is an increasing function of $t$?
It seems intuitive but I don't see how to prove it.
context:
Suppose we have some sets $A_1,...A_n$. let's choose a random subset $X$ of their union as follows: for each $x$ in their union, place $x$ in $X$ with probability $t$.
Therefore, for each subset $C$ of their union, $C$ is a subset of $X$ with probability $t^{|C|}$.
Consider the event $E$ = ($A_1$ is a subset of $X$) $cupdotscup$ ($A_n$ is a subset of $X$).
I want to show that $p(E)$ is an increasing function of $t$ (on the segment [0,1])
I apologize for the way this is written, I am writing from my phone
probability
edited Nov 21 '18 at 19:39
Bernard
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
|
show 3 more comments
Suppose we define a probability on some sample space, dependent on a parameter $t$. Let $E$ and $F$ be two events. Suppose that for each of these events, their probability is an increasing function of $t$.
Does this mean that the probability of the union of $E$ and $F$ is an increasing function of $t$?
It seems intuitive but I don't see how to prove it.
context:
Suppose we have some sets $A_1,...A_n$. let's choose a random subset $X$ of their union as follows: for each $x$ in their union, place $x$ in $X$ with probability $t$.
Therefore, for each subset $C$ of their union, $C$ is a subset of $X$ with probability $t^{|C|}$.
Consider the event $E$ = ($A_1$ is a subset of $X$) $cupdotscup$ ($A_n$ is a subset of $X$).
I want to show that $p(E)$ is an increasing function of $t$ (on the segment [0,1])
I apologize for the way this is written, I am writing from my phone
probability
edited Nov 21 '18 at 19:39
Bernard
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15
|
show 3 more comments
Suppose we define a probability on some sample space, dependent on a parameter $t$. Let $E$ and $F$ be two events. Suppose that for each of these events, their probability is an increasing function of $t$.
Does this mean that the probability of the union of $E$ and $F$ is an increasing function of $t$?
It seems intuitive but I don't see how to prove it.
context:
Suppose we have some sets $A_1,...A_n$. let's choose a random subset $X$ of their union as follows: for each $x$ in their union, place $x$ in $X$ with probability $t$.
Therefore, for each subset $C$ of their union, $C$ is a subset of $X$ with probability $t^{|C|}$.
Consider the event $E$ = ($A_1$ is a subset of $X$) $cupdotscup$ ($A_n$ is a subset of $X$).
I want to show that $p(E)$ is an increasing function of $t$ (on the segment [0,1])
I apologize for the way this is written, I am writing from my phone
probability
edited Nov 21 '18 at 19:39
Bernard
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
Suppose we define a probability on some sample space, dependent on a parameter $t$. Let $E$ and $F$ be two events. Suppose that for each of these events, their probability is an increasing function of $t$.
Does this mean that the probability of the union of $E$ and $F$ is an increasing function of $t$?
It seems intuitive but I don't see how to prove it.
context:
Suppose we have some sets $A_1,...A_n$. let's choose a random subset $X$ of their union as follows: for each $x$ in their union, place $x$ in $X$ with probability $t$.
Therefore, for each subset $C$ of their union, $C$ is a subset of $X$ with probability $t^{|C|}$.
Consider the event $E$ = ($A_1$ is a subset of $X$) $cupdotscup$ ($A_n$ is a subset of $X$).
I want to show that $p(E)$ is an increasing function of $t$ (on the segment [0,1])
I apologize for the way this is written, I am writing from my phone
probability
probability
edited Nov 21 '18 at 19:39
Bernard
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
edited Nov 21 '18 at 19:39
Bernard
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
edited Nov 21 '18 at 19:39
Bernard
118k639112
edited Nov 21 '18 at 19:39
Bernard
118k639112
edited Nov 21 '18 at 19:39
Bernard
118k639112
118k639112
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
asked Nov 21 '18 at 18:25
Joshua BenabouJoshua Benabou
2,552625
2,552625
Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15
|
show 3 more comments
Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15
Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15
|
show 3 more comments
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Hi, you are confused in more than one regard, it seems. Can you specify the context.
– Will M.
Nov 21 '18 at 18:31
@Will M. context added
– Joshua Benabou
Nov 21 '18 at 18:43
You should be careful with your formulation. Just asking that the probability of $E_t$ and $F_t$ is increasing is not enough. A (very artificial) counterexample might be considering $X$ a uniform on $[0,1]$, $E_t = {X < t}$ and $F_t = E_{1-t}^c$ for $t in [0, 1/2]$ and $F_t = E_t$ for $t in (1/2, 1]$.
– Daniel
Nov 21 '18 at 19:31
I am pretty sure I have seen the question on the context here on math stackexchange, but I can't find it.... However, the solution was about using inclusion exclusion principle and deriving the expression obtained (essentially we are deriving a polynomial)
– Daniel
Nov 21 '18 at 19:35
@Daniel: actually the original problem was to prove that the polynomial expression given by the inclusion exclusion principle is an increasing function of t, and one possible method seems to be interpreting that expression probabilistically and continuing as i tried to above. another possible idea is to derive the expression and show that the result is positive, which i have managed to do only in the case where the A_i are mutuslly disjoint.
– Joshua Benabou
Nov 21 '18 at 21:15