Mixing
A contradiction hidden in the definition of the probability of the intersection of events?
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Consider a set $C$ composed of three distinguishable kinds of elements $A,B,G$, and let $alpha,beta,gamma>0$ be the numbers of elements of each kind, and $|C|=c=alpha+beta+gamma$.
We define the three events $L_n^A, L_n^B, L_n^G$ as to get, in $n$ trials with replacement, at least one element of kind $A$, at least one element of kind $B$, and at least one element of kind $G$.
The probabilities of these events are $$P(L_n^A)=1-left(fracc-alphacright)^n=1-left(fracbeta+gammacright)^n, $$
$$P(L_n^B)=1-left(fracc-betacright)^n=1-left(fracalpha+gammacright)^n,$$
$$
P(L_n^G)=1-left(fracc-gammacright)^n=1-left(fracalpha+betacright)^n.
$$
We evaluate $P(L_n^Acap L_n^B)$. By definition of conditional probability and applying the property of the opposite event, we have
$$
P(L_n^Acap L_n^B)=P(L_n^A|L_n^B)P(L_n^B)=[1-P(overlineL_n^A|L_n^B)]P(L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B).
$$
By means of Bayes' theorem, $P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B|overlineL_n^A)P(overlineL_n^A)$. If we know that the event $L_n^A$ did not take place, all the $n$ extractions are either of kind $B$ or of kind $G$. Therefore, $P(L_n^B|overlineL_n^A)=1-left(fracbeta+gamma-betabeta+gammaright)^n=1-left(fracgammabeta+gammaright)^n$.
In conclusion, since $P(overlineL_n^A)=left(fracbeta+gammacright)^n$, we obtain
$$
P(L_n^Acap L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B)-P(L_n^B|overlineL_n^A)P(overlineL_n^A)=
$$
$$
=1-left(fracalpha+gammacright)^n-left[1-left(fracgammabeta+gammaright)^nright]left(fracbeta+gammacright)^n=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n.
$$
We notice that if $n=0$ (or $n=1$), then we correctly have $P(L_n^Acap L_n^B)=0$.
We now evaluate $P(L_n^Acap L_n^Bcap L_n^G)$. As we have done before, we find
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
which implies that,
for $n=0$, it must also be $P(L_n^Acap L_n^Bcap L_n^G)=0$.
We go on with the calculation of $P(L_n^Acap L_n^Bcap L_n^G)$, and we apply (again) first the definition of opposite event
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=[1-P(overlineL_n^G|L_n^Acap L_n^B)]P(L_n^Acap L_n^B)=
$$
$$
=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
and then the theorem of Bayes on the second term
$$
P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G).
$$
If we know that event $L_n^G$ does not take place, the probability to get at least one element of kind $A$ and at least one element of kind $B$ is
$$
P(L_n^Acap L_n^B|overlineL_n^G)=1-left(fracalphaalpha+betaright)^n-left(fracbetaalpha+betaright)^n.
$$
Therefore, since $P(overlineL_n^G)=left(fracalpha+betacright)^n$,
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B)-P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G)=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n-left(fracalpha+betacright)^n+left(fracalphacright)^n+left(fracbetacright)^n.
$$
If we now substitute $n=0$ in this expression we obtain $P(L_n^Acap L_n^Bcap L_n^G)=1$,
which is in contradiction with what we observed before, i.e. that $P(L_n^Acap L_n^Bcap L_n^G)=0$ with $n=0$.
There is likely a mistake in this reasoning, but I am not able to spot it.
Thanks for your help!
probability combinatorics
asked Jul 28 at 17:57
Andrea Prunotto
569114
add a comment |Â
up vote
1
down vote
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Consider a set $C$ composed of three distinguishable kinds of elements $A,B,G$, and let $alpha,beta,gamma>0$ be the numbers of elements of each kind, and $|C|=c=alpha+beta+gamma$.
We define the three events $L_n^A, L_n^B, L_n^G$ as to get, in $n$ trials with replacement, at least one element of kind $A$, at least one element of kind $B$, and at least one element of kind $G$.
The probabilities of these events are $$P(L_n^A)=1-left(fracc-alphacright)^n=1-left(fracbeta+gammacright)^n, $$
$$P(L_n^B)=1-left(fracc-betacright)^n=1-left(fracalpha+gammacright)^n,$$
$$
P(L_n^G)=1-left(fracc-gammacright)^n=1-left(fracalpha+betacright)^n.
$$
We evaluate $P(L_n^Acap L_n^B)$. By definition of conditional probability and applying the property of the opposite event, we have
$$
P(L_n^Acap L_n^B)=P(L_n^A|L_n^B)P(L_n^B)=[1-P(overlineL_n^A|L_n^B)]P(L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B).
$$
By means of Bayes' theorem, $P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B|overlineL_n^A)P(overlineL_n^A)$. If we know that the event $L_n^A$ did not take place, all the $n$ extractions are either of kind $B$ or of kind $G$. Therefore, $P(L_n^B|overlineL_n^A)=1-left(fracbeta+gamma-betabeta+gammaright)^n=1-left(fracgammabeta+gammaright)^n$.
In conclusion, since $P(overlineL_n^A)=left(fracbeta+gammacright)^n$, we obtain
$$
P(L_n^Acap L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B)-P(L_n^B|overlineL_n^A)P(overlineL_n^A)=
$$
$$
=1-left(fracalpha+gammacright)^n-left[1-left(fracgammabeta+gammaright)^nright]left(fracbeta+gammacright)^n=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n.
$$
We notice that if $n=0$ (or $n=1$), then we correctly have $P(L_n^Acap L_n^B)=0$.
We now evaluate $P(L_n^Acap L_n^Bcap L_n^G)$. As we have done before, we find
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
which implies that,
for $n=0$, it must also be $P(L_n^Acap L_n^Bcap L_n^G)=0$.
We go on with the calculation of $P(L_n^Acap L_n^Bcap L_n^G)$, and we apply (again) first the definition of opposite event
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=[1-P(overlineL_n^G|L_n^Acap L_n^B)]P(L_n^Acap L_n^B)=
$$
$$
=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
and then the theorem of Bayes on the second term
$$
P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G).
$$
If we know that event $L_n^G$ does not take place, the probability to get at least one element of kind $A$ and at least one element of kind $B$ is
$$
P(L_n^Acap L_n^B|overlineL_n^G)=1-left(fracalphaalpha+betaright)^n-left(fracbetaalpha+betaright)^n.
$$
Therefore, since $P(overlineL_n^G)=left(fracalpha+betacright)^n$,
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B)-P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G)=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n-left(fracalpha+betacright)^n+left(fracalphacright)^n+left(fracbetacright)^n.
$$
If we now substitute $n=0$ in this expression we obtain $P(L_n^Acap L_n^Bcap L_n^G)=1$,
which is in contradiction with what we observed before, i.e. that $P(L_n^Acap L_n^Bcap L_n^G)=0$ with $n=0$.
There is likely a mistake in this reasoning, but I am not able to spot it.
Thanks for your help!
probability combinatorics
asked Jul 28 at 17:57
Andrea Prunotto
569114
2
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider a set $C$ composed of three distinguishable kinds of elements $A,B,G$, and let $alpha,beta,gamma>0$ be the numbers of elements of each kind, and $|C|=c=alpha+beta+gamma$.
We define the three events $L_n^A, L_n^B, L_n^G$ as to get, in $n$ trials with replacement, at least one element of kind $A$, at least one element of kind $B$, and at least one element of kind $G$.
The probabilities of these events are $$P(L_n^A)=1-left(fracc-alphacright)^n=1-left(fracbeta+gammacright)^n, $$
$$P(L_n^B)=1-left(fracc-betacright)^n=1-left(fracalpha+gammacright)^n,$$
$$
P(L_n^G)=1-left(fracc-gammacright)^n=1-left(fracalpha+betacright)^n.
$$
We evaluate $P(L_n^Acap L_n^B)$. By definition of conditional probability and applying the property of the opposite event, we have
$$
P(L_n^Acap L_n^B)=P(L_n^A|L_n^B)P(L_n^B)=[1-P(overlineL_n^A|L_n^B)]P(L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B).
$$
By means of Bayes' theorem, $P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B|overlineL_n^A)P(overlineL_n^A)$. If we know that the event $L_n^A$ did not take place, all the $n$ extractions are either of kind $B$ or of kind $G$. Therefore, $P(L_n^B|overlineL_n^A)=1-left(fracbeta+gamma-betabeta+gammaright)^n=1-left(fracgammabeta+gammaright)^n$.
In conclusion, since $P(overlineL_n^A)=left(fracbeta+gammacright)^n$, we obtain
$$
P(L_n^Acap L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B)-P(L_n^B|overlineL_n^A)P(overlineL_n^A)=
$$
$$
=1-left(fracalpha+gammacright)^n-left[1-left(fracgammabeta+gammaright)^nright]left(fracbeta+gammacright)^n=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n.
$$
We notice that if $n=0$ (or $n=1$), then we correctly have $P(L_n^Acap L_n^B)=0$.
We now evaluate $P(L_n^Acap L_n^Bcap L_n^G)$. As we have done before, we find
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
which implies that,
for $n=0$, it must also be $P(L_n^Acap L_n^Bcap L_n^G)=0$.
We go on with the calculation of $P(L_n^Acap L_n^Bcap L_n^G)$, and we apply (again) first the definition of opposite event
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=[1-P(overlineL_n^G|L_n^Acap L_n^B)]P(L_n^Acap L_n^B)=
$$
$$
=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
and then the theorem of Bayes on the second term
$$
P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G).
$$
If we know that event $L_n^G$ does not take place, the probability to get at least one element of kind $A$ and at least one element of kind $B$ is
$$
P(L_n^Acap L_n^B|overlineL_n^G)=1-left(fracalphaalpha+betaright)^n-left(fracbetaalpha+betaright)^n.
$$
Therefore, since $P(overlineL_n^G)=left(fracalpha+betacright)^n$,
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B)-P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G)=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n-left(fracalpha+betacright)^n+left(fracalphacright)^n+left(fracbetacright)^n.
$$
If we now substitute $n=0$ in this expression we obtain $P(L_n^Acap L_n^Bcap L_n^G)=1$,
which is in contradiction with what we observed before, i.e. that $P(L_n^Acap L_n^Bcap L_n^G)=0$ with $n=0$.
There is likely a mistake in this reasoning, but I am not able to spot it.
Thanks for your help!
probability combinatorics
asked Jul 28 at 17:57
Andrea Prunotto
569114
Consider a set $C$ composed of three distinguishable kinds of elements $A,B,G$, and let $alpha,beta,gamma>0$ be the numbers of elements of each kind, and $|C|=c=alpha+beta+gamma$.
We define the three events $L_n^A, L_n^B, L_n^G$ as to get, in $n$ trials with replacement, at least one element of kind $A$, at least one element of kind $B$, and at least one element of kind $G$.
The probabilities of these events are $$P(L_n^A)=1-left(fracc-alphacright)^n=1-left(fracbeta+gammacright)^n, $$
$$P(L_n^B)=1-left(fracc-betacright)^n=1-left(fracalpha+gammacright)^n,$$
$$
P(L_n^G)=1-left(fracc-gammacright)^n=1-left(fracalpha+betacright)^n.
$$
We evaluate $P(L_n^Acap L_n^B)$. By definition of conditional probability and applying the property of the opposite event, we have
$$
P(L_n^Acap L_n^B)=P(L_n^A|L_n^B)P(L_n^B)=[1-P(overlineL_n^A|L_n^B)]P(L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B).
$$
By means of Bayes' theorem, $P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B|overlineL_n^A)P(overlineL_n^A)$. If we know that the event $L_n^A$ did not take place, all the $n$ extractions are either of kind $B$ or of kind $G$. Therefore, $P(L_n^B|overlineL_n^A)=1-left(fracbeta+gamma-betabeta+gammaright)^n=1-left(fracgammabeta+gammaright)^n$.
In conclusion, since $P(overlineL_n^A)=left(fracbeta+gammacright)^n$, we obtain
$$
P(L_n^Acap L_n^B)=P(L_n^B)-P(overlineL_n^A|L_n^B)P(L_n^B)=P(L_n^B)-P(L_n^B|overlineL_n^A)P(overlineL_n^A)=
$$
$$
=1-left(fracalpha+gammacright)^n-left[1-left(fracgammabeta+gammaright)^nright]left(fracbeta+gammacright)^n=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n.
$$
We notice that if $n=0$ (or $n=1$), then we correctly have $P(L_n^Acap L_n^B)=0$.
We now evaluate $P(L_n^Acap L_n^Bcap L_n^G)$. As we have done before, we find
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
which implies that,
for $n=0$, it must also be $P(L_n^Acap L_n^Bcap L_n^G)=0$.
We go on with the calculation of $P(L_n^Acap L_n^Bcap L_n^G)$, and we apply (again) first the definition of opposite event
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=[1-P(overlineL_n^G|L_n^Acap L_n^B)]P(L_n^Acap L_n^B)=
$$
$$
=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B),
$$
and then the theorem of Bayes on the second term
$$
P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G).
$$
If we know that event $L_n^G$ does not take place, the probability to get at least one element of kind $A$ and at least one element of kind $B$ is
$$
P(L_n^Acap L_n^B|overlineL_n^G)=1-left(fracalphaalpha+betaright)^n-left(fracbetaalpha+betaright)^n.
$$
Therefore, since $P(overlineL_n^G)=left(fracalpha+betacright)^n$,
$$
P(L_n^Acap L_n^Bcap L_n^G)=P(L_n^Acap L_n^B)-P(overlineL_n^G|L_n^Acap L_n^B)P(L_n^Acap L_n^B)=P(L_n^Acap L_n^B)-P(L_n^Acap L_n^B|overlineL_n^G)P(overlineL_n^G)=
$$
$$
=1-left(fracalpha+gammacright)^n-left(fracbeta+gammacright)^n+left(fracgammacright)^n-left(fracalpha+betacright)^n+left(fracalphacright)^n+left(fracbetacright)^n.
$$
If we now substitute $n=0$ in this expression we obtain $P(L_n^Acap L_n^Bcap L_n^G)=1$,
which is in contradiction with what we observed before, i.e. that $P(L_n^Acap L_n^Bcap L_n^G)=0$ with $n=0$.
There is likely a mistake in this reasoning, but I am not able to spot it.
Thanks for your help!
probability combinatorics
asked Jul 28 at 17:57
Andrea Prunotto
569114
asked Jul 28 at 17:57
Andrea Prunotto
569114
asked Jul 28 at 17:57
Andrea Prunotto
569114
asked Jul 28 at 17:57
Andrea Prunotto
569114
569114
2
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03
add a comment |Â
2
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03
2
2
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03
add a comment |Â
1 Answer
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There are a couple of problems that I can see.
- You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
- In some cases your formulae are only valid for $ngeq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $ngeq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
answered Jul 28 at 18:22
Especially Lime
19.1k22252
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
add a comment |Â
1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There are a couple of problems that I can see.
- You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
- In some cases your formulae are only valid for $ngeq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $ngeq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
answered Jul 28 at 18:22
Especially Lime
19.1k22252
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
add a comment |Â
up vote
1
down vote
accepted
There are a couple of problems that I can see.
- You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
- In some cases your formulae are only valid for $ngeq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $ngeq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
answered Jul 28 at 18:22
Especially Lime
19.1k22252
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There are a couple of problems that I can see.
- You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
- In some cases your formulae are only valid for $ngeq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $ngeq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
answered Jul 28 at 18:22
Especially Lime
19.1k22252
There are a couple of problems that I can see.
- You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
- In some cases your formulae are only valid for $ngeq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $ngeq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
answered Jul 28 at 18:22
Especially Lime
19.1k22252
answered Jul 28 at 18:22
Especially Lime
19.1k22252
answered Jul 28 at 18:22
Especially Lime
19.1k22252
answered Jul 28 at 18:22
Especially Lime
19.1k22252
19.1k22252
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
add a comment |Â
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
Thanks! I see very well now!
– Andrea Prunotto
Jul 28 at 18:28
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^Acap L_n^B)$ but not for $P(L_n^Acap L_n^Bcap L_n^G)$, although we used in both cases Bayes' theorem...
– Andrea Prunotto
Jul 28 at 18:36
1
1
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation.
– Especially Lime
Jul 28 at 18:56
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
But then it is however $P(L_0^Acap L_0^Bcap L_0^G)=P(L_0^Acap L_0^B)=0$, isn't it? How can I prove it?
– Andrea Prunotto
Jul 28 at 19:25
add a comment |Â
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2
What’s the Cliffs notes version?
– Randall
Jul 28 at 18:01
@Randall Hi Randall, sorry I don't know what are you referring to!
– Andrea Prunotto
Jul 28 at 18:03