Estimate the relationship between the probabilities of head of two biased coins
Question: Assume that $p$ and $q$ were uniformly sampled in $[0, 1]$ and two biased coins whose probabilities of head are $p$ and $q$, respectively, were made. However, we don't know what $p$ and $q$ are. To estimate them, each coin is tossed $n$ times, then we got $r_1$ heads for the coin whose probability of head is $p$ and $r_2$ heads for the other coin as a result. What is the probability that $p<q$?
I tried Bayes theorem to solve above question. Let $A$ and $B$ denote
$A$: An event that $p<q$
$B$: An event that we got the results of tosses as mentioned above
Then
$$ P(A) = frac{1}{2} quad text{and} quad P(B) = binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2} $$
trivially, and what I want to calculate is $P(A|B)$. However, I don't have any idea to get $P(B|A)$ to use Bayes theorem. How can I solve this?
probability conditional-probability bayes-theorem
asked Dec 4 at 5:57
Jeongu Kim
734
add a comment |
Question: Assume that $p$ and $q$ were uniformly sampled in $[0, 1]$ and two biased coins whose probabilities of head are $p$ and $q$, respectively, were made. However, we don't know what $p$ and $q$ are. To estimate them, each coin is tossed $n$ times, then we got $r_1$ heads for the coin whose probability of head is $p$ and $r_2$ heads for the other coin as a result. What is the probability that $p<q$?
I tried Bayes theorem to solve above question. Let $A$ and $B$ denote
$A$: An event that $p<q$
$B$: An event that we got the results of tosses as mentioned above
Then
$$ P(A) = frac{1}{2} quad text{and} quad P(B) = binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2} $$
trivially, and what I want to calculate is $P(A|B)$. However, I don't have any idea to get $P(B|A)$ to use Bayes theorem. How can I solve this?
probability conditional-probability bayes-theorem
asked Dec 4 at 5:57
Jeongu Kim
734
add a comment |
Question: Assume that $p$ and $q$ were uniformly sampled in $[0, 1]$ and two biased coins whose probabilities of head are $p$ and $q$, respectively, were made. However, we don't know what $p$ and $q$ are. To estimate them, each coin is tossed $n$ times, then we got $r_1$ heads for the coin whose probability of head is $p$ and $r_2$ heads for the other coin as a result. What is the probability that $p<q$?
I tried Bayes theorem to solve above question. Let $A$ and $B$ denote
$A$: An event that $p<q$
$B$: An event that we got the results of tosses as mentioned above
Then
$$ P(A) = frac{1}{2} quad text{and} quad P(B) = binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2} $$
trivially, and what I want to calculate is $P(A|B)$. However, I don't have any idea to get $P(B|A)$ to use Bayes theorem. How can I solve this?
probability conditional-probability bayes-theorem
asked Dec 4 at 5:57
Jeongu Kim
734
Question: Assume that $p$ and $q$ were uniformly sampled in $[0, 1]$ and two biased coins whose probabilities of head are $p$ and $q$, respectively, were made. However, we don't know what $p$ and $q$ are. To estimate them, each coin is tossed $n$ times, then we got $r_1$ heads for the coin whose probability of head is $p$ and $r_2$ heads for the other coin as a result. What is the probability that $p<q$?
I tried Bayes theorem to solve above question. Let $A$ and $B$ denote
$A$: An event that $p<q$
$B$: An event that we got the results of tosses as mentioned above
Then
$$ P(A) = frac{1}{2} quad text{and} quad P(B) = binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2} $$
trivially, and what I want to calculate is $P(A|B)$. However, I don't have any idea to get $P(B|A)$ to use Bayes theorem. How can I solve this?
probability conditional-probability bayes-theorem
probability conditional-probability bayes-theorem
asked Dec 4 at 5:57
Jeongu Kim
734
asked Dec 4 at 5:57
Jeongu Kim
734
edited Dec 7 at 7:39
asked Dec 4 at 5:57
Jeongu Kim
734
asked Dec 4 at 5:57
Jeongu Kim
734
asked Dec 4 at 5:57
Jeongu Kim
734
734
add a comment |
add a comment |
1 Answer
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You are correct that $P(A)=1/2$, but your claimed formula for $P(B)$ is in fact the formula for $P(B|p,q)$, i.e. the probability of the event $B$ given that you know $p$ and $q$. Baye's theorem states that $P(A|B) = P(B|A) cdot P(A)/P(B)$. So you are left with computing $P(B|A)$ and $P(B)$. This boils down to computing the integral of $binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2}$ over $p$ and $q$ in the appropriate ranges. You will probably find the Beta function helpful.
answered Dec 4 at 8:12
norfair
99512
add a comment |
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1 Answer
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1 Answer
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You are correct that $P(A)=1/2$, but your claimed formula for $P(B)$ is in fact the formula for $P(B|p,q)$, i.e. the probability of the event $B$ given that you know $p$ and $q$. Baye's theorem states that $P(A|B) = P(B|A) cdot P(A)/P(B)$. So you are left with computing $P(B|A)$ and $P(B)$. This boils down to computing the integral of $binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2}$ over $p$ and $q$ in the appropriate ranges. You will probably find the Beta function helpful.
answered Dec 4 at 8:12
norfair
99512
add a comment |
You are correct that $P(A)=1/2$, but your claimed formula for $P(B)$ is in fact the formula for $P(B|p,q)$, i.e. the probability of the event $B$ given that you know $p$ and $q$. Baye's theorem states that $P(A|B) = P(B|A) cdot P(A)/P(B)$. So you are left with computing $P(B|A)$ and $P(B)$. This boils down to computing the integral of $binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2}$ over $p$ and $q$ in the appropriate ranges. You will probably find the Beta function helpful.
answered Dec 4 at 8:12
norfair
99512
add a comment |
You are correct that $P(A)=1/2$, but your claimed formula for $P(B)$ is in fact the formula for $P(B|p,q)$, i.e. the probability of the event $B$ given that you know $p$ and $q$. Baye's theorem states that $P(A|B) = P(B|A) cdot P(A)/P(B)$. So you are left with computing $P(B|A)$ and $P(B)$. This boils down to computing the integral of $binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2}$ over $p$ and $q$ in the appropriate ranges. You will probably find the Beta function helpful.
answered Dec 4 at 8:12
norfair
99512
You are correct that $P(A)=1/2$, but your claimed formula for $P(B)$ is in fact the formula for $P(B|p,q)$, i.e. the probability of the event $B$ given that you know $p$ and $q$. Baye's theorem states that $P(A|B) = P(B|A) cdot P(A)/P(B)$. So you are left with computing $P(B|A)$ and $P(B)$. This boils down to computing the integral of $binom{n}{r_1}p^{r_1}(1-p)^{n-r_1}binom{n}{r_2}q^{r_2}(1-q)^{n-r_2}$ over $p$ and $q$ in the appropriate ranges. You will probably find the Beta function helpful.
answered Dec 4 at 8:12
norfair
99512
answered Dec 4 at 8:12
norfair
99512
answered Dec 4 at 8:12
norfair
99512
answered Dec 4 at 8:12
norfair
99512
99512
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