Mixing
The total probability as sums of conditional probabilities
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ and $C_2'$ partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements (denoted as $C_1'$ and $C_2'$) are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
This, in turn, is (I think) given by,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked/ failed to invoke.
probability probability-theory
asked Nov 20 '18 at 22:57
user617643
226
|
show 1 more comment
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ and $C_2'$ partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements (denoted as $C_1'$ and $C_2'$) are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
This, in turn, is (I think) given by,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked/ failed to invoke.
probability probability-theory
asked Nov 20 '18 at 22:57
user617643
226
Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35
|
show 1 more comment
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ and $C_2'$ partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements (denoted as $C_1'$ and $C_2'$) are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
This, in turn, is (I think) given by,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked/ failed to invoke.
probability probability-theory
asked Nov 20 '18 at 22:57
user617643
226
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ and $C_2'$ partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements (denoted as $C_1'$ and $C_2'$) are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
This, in turn, is (I think) given by,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked/ failed to invoke.
probability probability-theory
probability probability-theory
asked Nov 20 '18 at 22:57
user617643
226
asked Nov 20 '18 at 22:57
user617643
226
edited Nov 20 '18 at 23:33
asked Nov 20 '18 at 22:57
user617643
226
asked Nov 20 '18 at 22:57
user617643
226
asked Nov 20 '18 at 22:57
user617643
226
226
Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35
|
show 1 more comment
Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35
Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35
|
show 1 more comment
1 Answer
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Firstly the partition you require is ${C_1cap C_2, C_1cap C_2', C_1'cap C_2,C_1'cap C_2'}$, the pairwise intersections of the two events or their complements (excluding the empty intersections; assuming all four shown are non-empty).
Secondly, take more care when expanding conditionals; let's do it one step at a time.
$begin{align}mathsf P(X) &= mathsf P(X,C_1,C_2)+mathsf P(X,C_1,C_2')+mathsf P(X,C_1',C_2)+mathsf P(X,C_1',C_2')\[2ex] &= {mathsf P(C_1, C_2)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1,C_2')mathsf P(Xmid C_1,C_2')+mathsf P(C_1',C_2)mathsf P(Xmid C_1',C_2)+mathsf P(C_1',C_2'mid C_1')mathsf P(Xmid C_1',C_2')}\[2ex] &= {mathsf P(C_1)mathsf P(C_2mid C_1)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1)mathsf P(C_2'mid C_1)mathsf P(Xmid C_1,C_2')+mathsf P(C_1')mathsf P(C_2mid C_1')mathsf P(Xmid C_1',C_2)+mathsf P(C_1')mathsf P(C_2'mid C_1')mathsf P(Xmid C_1',C_2')}end{align}$
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
add a comment |
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Firstly the partition you require is ${C_1cap C_2, C_1cap C_2', C_1'cap C_2,C_1'cap C_2'}$, the pairwise intersections of the two events or their complements (excluding the empty intersections; assuming all four shown are non-empty).
Secondly, take more care when expanding conditionals; let's do it one step at a time.
$begin{align}mathsf P(X) &= mathsf P(X,C_1,C_2)+mathsf P(X,C_1,C_2')+mathsf P(X,C_1',C_2)+mathsf P(X,C_1',C_2')\[2ex] &= {mathsf P(C_1, C_2)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1,C_2')mathsf P(Xmid C_1,C_2')+mathsf P(C_1',C_2)mathsf P(Xmid C_1',C_2)+mathsf P(C_1',C_2'mid C_1')mathsf P(Xmid C_1',C_2')}\[2ex] &= {mathsf P(C_1)mathsf P(C_2mid C_1)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1)mathsf P(C_2'mid C_1)mathsf P(Xmid C_1,C_2')+mathsf P(C_1')mathsf P(C_2mid C_1')mathsf P(Xmid C_1',C_2)+mathsf P(C_1')mathsf P(C_2'mid C_1')mathsf P(Xmid C_1',C_2')}end{align}$
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
add a comment |
Firstly the partition you require is ${C_1cap C_2, C_1cap C_2', C_1'cap C_2,C_1'cap C_2'}$, the pairwise intersections of the two events or their complements (excluding the empty intersections; assuming all four shown are non-empty).
Secondly, take more care when expanding conditionals; let's do it one step at a time.
$begin{align}mathsf P(X) &= mathsf P(X,C_1,C_2)+mathsf P(X,C_1,C_2')+mathsf P(X,C_1',C_2)+mathsf P(X,C_1',C_2')\[2ex] &= {mathsf P(C_1, C_2)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1,C_2')mathsf P(Xmid C_1,C_2')+mathsf P(C_1',C_2)mathsf P(Xmid C_1',C_2)+mathsf P(C_1',C_2'mid C_1')mathsf P(Xmid C_1',C_2')}\[2ex] &= {mathsf P(C_1)mathsf P(C_2mid C_1)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1)mathsf P(C_2'mid C_1)mathsf P(Xmid C_1,C_2')+mathsf P(C_1')mathsf P(C_2mid C_1')mathsf P(Xmid C_1',C_2)+mathsf P(C_1')mathsf P(C_2'mid C_1')mathsf P(Xmid C_1',C_2')}end{align}$
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
add a comment |
Firstly the partition you require is ${C_1cap C_2, C_1cap C_2', C_1'cap C_2,C_1'cap C_2'}$, the pairwise intersections of the two events or their complements (excluding the empty intersections; assuming all four shown are non-empty).
Secondly, take more care when expanding conditionals; let's do it one step at a time.
$begin{align}mathsf P(X) &= mathsf P(X,C_1,C_2)+mathsf P(X,C_1,C_2')+mathsf P(X,C_1',C_2)+mathsf P(X,C_1',C_2')\[2ex] &= {mathsf P(C_1, C_2)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1,C_2')mathsf P(Xmid C_1,C_2')+mathsf P(C_1',C_2)mathsf P(Xmid C_1',C_2)+mathsf P(C_1',C_2'mid C_1')mathsf P(Xmid C_1',C_2')}\[2ex] &= {mathsf P(C_1)mathsf P(C_2mid C_1)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1)mathsf P(C_2'mid C_1)mathsf P(Xmid C_1,C_2')+mathsf P(C_1')mathsf P(C_2mid C_1')mathsf P(Xmid C_1',C_2)+mathsf P(C_1')mathsf P(C_2'mid C_1')mathsf P(Xmid C_1',C_2')}end{align}$
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
Firstly the partition you require is ${C_1cap C_2, C_1cap C_2', C_1'cap C_2,C_1'cap C_2'}$, the pairwise intersections of the two events or their complements (excluding the empty intersections; assuming all four shown are non-empty).
Secondly, take more care when expanding conditionals; let's do it one step at a time.
$begin{align}mathsf P(X) &= mathsf P(X,C_1,C_2)+mathsf P(X,C_1,C_2')+mathsf P(X,C_1',C_2)+mathsf P(X,C_1',C_2')\[2ex] &= {mathsf P(C_1, C_2)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1,C_2')mathsf P(Xmid C_1,C_2')+mathsf P(C_1',C_2)mathsf P(Xmid C_1',C_2)+mathsf P(C_1',C_2'mid C_1')mathsf P(Xmid C_1',C_2')}\[2ex] &= {mathsf P(C_1)mathsf P(C_2mid C_1)mathsf P(Xmid C_1,C_2)+ mathsf P(C_1)mathsf P(C_2'mid C_1)mathsf P(Xmid C_1,C_2')+mathsf P(C_1')mathsf P(C_2mid C_1')mathsf P(Xmid C_1',C_2)+mathsf P(C_1')mathsf P(C_2'mid C_1')mathsf P(Xmid C_1',C_2')}end{align}$
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
edited Nov 20 '18 at 23:51
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
answered Nov 20 '18 at 23:41
Graham Kemp
84.7k43378
84.7k43378
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
add a comment |
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
Thanks Graham. Could you take a look at my other question, where I elaborate on how I got my expression.
– user617643
Nov 21 '18 at 19:55
add a comment |
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Are you denoting complements with $´$?
– Will M.
Nov 20 '18 at 23:18
Yes I am! Will clarify this.
– user617643
Nov 20 '18 at 23:19
Something is fishy, if $E$ is any event, then any other no null event $F$ will be such that either $E cap F$ or else $E' cap F$ is not null. So, $C_1, C_2, C_1'$ and $C_2'$ cannot be a partition.
– Will M.
Nov 20 '18 at 23:23
I'm not sure I understand what you mean.
– user617643
Nov 20 '18 at 23:33
You can't have two events such that these two and their complements form a partition of the space. That is impossible.
– Will M.
Nov 20 '18 at 23:35