Mixing
Can you explain this counterintuitive conditional expectation result intuitively?
$begingroup$
Consider the following experiment.
We throw a three-sided die with sides $1$, $2$ and $3$ infinitely many times. Let $T_i$ denote the outcome of the $i$'th throw. Define $N:=min{i:T_ineq1}$. Let $X$ be the event that $T_N=2$ and let $Y$ be the event that $T_N=3$.
Some calculation (*) leads to the result that $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
Let $Z$ be the event that $T_ineq3$ for all $i$. Some calculation (**) leads to the result that $mathbb{E}(N|Z)=2$.
I find it very unintuitive that $mathbb{E}(N|X)neqmathbb{E}(N|Z)$. Obviously we have $Xsubsetneq Z$. However, the information $Z$ gives, which $X$ does not give, intuitively only affects what comes after the $N$'th throw. So how is it possible that the probability distribution of $N$ is different when conditioning on $X$ or $Z$?
(*) We have $mathbb{P}(X)=mathbb{P}(Y)$ and $mathbb{E}(N|X)=mathbb{E}(N|Y)$ by symmetry. Also notice that $X$ and $Y$ partition the event space, so $mathbb{P}(X)+mathbb{P}(Y)=1$, so $mathbb{P}(X)=mathbb{P}(Y)=frac12$. Since $mathbb{P}(T_ineq1)=2/3$, we have $mathbb{E}(N)=3/2$. By the principle of divide and conquer, we have $mathbb{E}(N)=mathbb{P}(X)mathbb{E}(N|X)+mathbb{P}(Y)mathbb{E}(N|Y)$, so we find $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
(**) We have $mathbb{P}(T_ineq1|Z)=1/2$, so $mathbb{E}(N|Z)=2/1=2$.
By the way, if you have a suggestion for a better, more specific title, be my guest. I could not come up with a good descriptive title for this very specific question.
probability intuition conditional-expectation dice
edited Feb 10 at 10:52
jvdhooft
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
$endgroup$
add a comment |
$begingroup$
Consider the following experiment.
We throw a three-sided die with sides $1$, $2$ and $3$ infinitely many times. Let $T_i$ denote the outcome of the $i$'th throw. Define $N:=min{i:T_ineq1}$. Let $X$ be the event that $T_N=2$ and let $Y$ be the event that $T_N=3$.
Some calculation (*) leads to the result that $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
Let $Z$ be the event that $T_ineq3$ for all $i$. Some calculation (**) leads to the result that $mathbb{E}(N|Z)=2$.
I find it very unintuitive that $mathbb{E}(N|X)neqmathbb{E}(N|Z)$. Obviously we have $Xsubsetneq Z$. However, the information $Z$ gives, which $X$ does not give, intuitively only affects what comes after the $N$'th throw. So how is it possible that the probability distribution of $N$ is different when conditioning on $X$ or $Z$?
(*) We have $mathbb{P}(X)=mathbb{P}(Y)$ and $mathbb{E}(N|X)=mathbb{E}(N|Y)$ by symmetry. Also notice that $X$ and $Y$ partition the event space, so $mathbb{P}(X)+mathbb{P}(Y)=1$, so $mathbb{P}(X)=mathbb{P}(Y)=frac12$. Since $mathbb{P}(T_ineq1)=2/3$, we have $mathbb{E}(N)=3/2$. By the principle of divide and conquer, we have $mathbb{E}(N)=mathbb{P}(X)mathbb{E}(N|X)+mathbb{P}(Y)mathbb{E}(N|Y)$, so we find $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
(**) We have $mathbb{P}(T_ineq1|Z)=1/2$, so $mathbb{E}(N|Z)=2/1=2$.
By the way, if you have a suggestion for a better, more specific title, be my guest. I could not come up with a good descriptive title for this very specific question.
probability intuition conditional-expectation dice
edited Feb 10 at 10:52
jvdhooft
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
$endgroup$
add a comment |
$begingroup$
Consider the following experiment.
We throw a three-sided die with sides $1$, $2$ and $3$ infinitely many times. Let $T_i$ denote the outcome of the $i$'th throw. Define $N:=min{i:T_ineq1}$. Let $X$ be the event that $T_N=2$ and let $Y$ be the event that $T_N=3$.
Some calculation (*) leads to the result that $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
Let $Z$ be the event that $T_ineq3$ for all $i$. Some calculation (**) leads to the result that $mathbb{E}(N|Z)=2$.
I find it very unintuitive that $mathbb{E}(N|X)neqmathbb{E}(N|Z)$. Obviously we have $Xsubsetneq Z$. However, the information $Z$ gives, which $X$ does not give, intuitively only affects what comes after the $N$'th throw. So how is it possible that the probability distribution of $N$ is different when conditioning on $X$ or $Z$?
(*) We have $mathbb{P}(X)=mathbb{P}(Y)$ and $mathbb{E}(N|X)=mathbb{E}(N|Y)$ by symmetry. Also notice that $X$ and $Y$ partition the event space, so $mathbb{P}(X)+mathbb{P}(Y)=1$, so $mathbb{P}(X)=mathbb{P}(Y)=frac12$. Since $mathbb{P}(T_ineq1)=2/3$, we have $mathbb{E}(N)=3/2$. By the principle of divide and conquer, we have $mathbb{E}(N)=mathbb{P}(X)mathbb{E}(N|X)+mathbb{P}(Y)mathbb{E}(N|Y)$, so we find $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
(**) We have $mathbb{P}(T_ineq1|Z)=1/2$, so $mathbb{E}(N|Z)=2/1=2$.
By the way, if you have a suggestion for a better, more specific title, be my guest. I could not come up with a good descriptive title for this very specific question.
probability intuition conditional-expectation dice
edited Feb 10 at 10:52
jvdhooft
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
$endgroup$
Consider the following experiment.
We throw a three-sided die with sides $1$, $2$ and $3$ infinitely many times. Let $T_i$ denote the outcome of the $i$'th throw. Define $N:=min{i:T_ineq1}$. Let $X$ be the event that $T_N=2$ and let $Y$ be the event that $T_N=3$.
Some calculation (*) leads to the result that $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
Let $Z$ be the event that $T_ineq3$ for all $i$. Some calculation (**) leads to the result that $mathbb{E}(N|Z)=2$.
I find it very unintuitive that $mathbb{E}(N|X)neqmathbb{E}(N|Z)$. Obviously we have $Xsubsetneq Z$. However, the information $Z$ gives, which $X$ does not give, intuitively only affects what comes after the $N$'th throw. So how is it possible that the probability distribution of $N$ is different when conditioning on $X$ or $Z$?
(*) We have $mathbb{P}(X)=mathbb{P}(Y)$ and $mathbb{E}(N|X)=mathbb{E}(N|Y)$ by symmetry. Also notice that $X$ and $Y$ partition the event space, so $mathbb{P}(X)+mathbb{P}(Y)=1$, so $mathbb{P}(X)=mathbb{P}(Y)=frac12$. Since $mathbb{P}(T_ineq1)=2/3$, we have $mathbb{E}(N)=3/2$. By the principle of divide and conquer, we have $mathbb{E}(N)=mathbb{P}(X)mathbb{E}(N|X)+mathbb{P}(Y)mathbb{E}(N|Y)$, so we find $mathbb{E}(N)=mathbb{E}(N|X)=mathbb{E}(N|Y)=3/2$.
(**) We have $mathbb{P}(T_ineq1|Z)=1/2$, so $mathbb{E}(N|Z)=2/1=2$.
By the way, if you have a suggestion for a better, more specific title, be my guest. I could not come up with a good descriptive title for this very specific question.
probability intuition conditional-expectation dice
probability intuition conditional-expectation dice
edited Feb 10 at 10:52
jvdhooft
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
edited Feb 10 at 10:52
jvdhooft
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
edited Feb 10 at 10:52
jvdhooft
5,67561641
edited Feb 10 at 10:52
jvdhooft
5,67561641
edited Feb 10 at 10:52
jvdhooft
5,67561641
5,67561641
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
asked Jan 28 at 22:46
SmileyCraftSmileyCraft
3,761519
3,761519
add a comment |
add a comment |
1 Answer
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$begingroup$
I think the main problem is that $P(Z)=0$, that means $Z$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $frac00$ limit forms in calculus, as both $P(N cap Z)=0$ and $P(Z)=0$.
answered Jan 28 at 23:13
IngixIngix
5,097159
$endgroup$
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think the main problem is that $P(Z)=0$, that means $Z$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $frac00$ limit forms in calculus, as both $P(N cap Z)=0$ and $P(Z)=0$.
answered Jan 28 at 23:13
IngixIngix
5,097159
$endgroup$
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
add a comment |
$begingroup$
I think the main problem is that $P(Z)=0$, that means $Z$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $frac00$ limit forms in calculus, as both $P(N cap Z)=0$ and $P(Z)=0$.
answered Jan 28 at 23:13
IngixIngix
5,097159
$endgroup$
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
add a comment |
$begingroup$
I think the main problem is that $P(Z)=0$, that means $Z$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $frac00$ limit forms in calculus, as both $P(N cap Z)=0$ and $P(Z)=0$.
answered Jan 28 at 23:13
IngixIngix
5,097159
$endgroup$
I think the main problem is that $P(Z)=0$, that means $Z$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $frac00$ limit forms in calculus, as both $P(N cap Z)=0$ and $P(Z)=0$.
answered Jan 28 at 23:13
IngixIngix
5,097159
answered Jan 28 at 23:13
IngixIngix
5,097159
answered Jan 28 at 23:13
IngixIngix
5,097159
answered Jan 28 at 23:13
IngixIngix
5,097159
5,097159
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
add a comment |
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
$begingroup$
In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_ineq1$ for all $ileq n$. Then for any event $omega$ we have $limmathbb{P}(omega|Z_n)=mathbb{P}(omega|Z)$. This way you can avoid conditioning on an event of zero probability.
$endgroup$
– SmileyCraft
Jan 28 at 23:23
add a comment |
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