The probability of heads of a random coin is uniform r.v. P. Find the probability that heads will show?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
The question states:
The probability of heads of a random coin is a random variable P, uniform in the interval $[0.4,0.6]$. Find the probability that at the next tossing of the coin that heads will show. Suppose the coin is tossed $50$ times resulting in $21$ tails and $29$ heads. What is $P(p|Observed Data)$?
My first question is what is the probability of P. It seems like there are two different ways to look at it. Is the probability such that $f(p)=5$ for $0.4 le p le 0.6$ and zero otherwise, thus
$$P(H) = int_0.4^0.6pdp = 0.5$$
or on the other hand does it follow the standard simple uniform distribution. That is
$$P[0.4 le P le 0.6] = int_0.4^0.6 dp = 0.2$$
Then, to answer the second part, is it correctly broken down by
$$fracp^29(1-p)^21int_0.4^0.6p^29(1-p)^21dp$$
for $0.4 le p le 0.6$ and $0$ otherwise. Thus, we obtain
$$P(H|A) = int_0.4^0.6 pf(p|A)dp$$
Thank you so much for your help in advance, I really appreciate it!
probability random-variables uniform-distribution
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
add a comment |Â
up vote
0
down vote
favorite
The question states:
The probability of heads of a random coin is a random variable P, uniform in the interval $[0.4,0.6]$. Find the probability that at the next tossing of the coin that heads will show. Suppose the coin is tossed $50$ times resulting in $21$ tails and $29$ heads. What is $P(p|Observed Data)$?
My first question is what is the probability of P. It seems like there are two different ways to look at it. Is the probability such that $f(p)=5$ for $0.4 le p le 0.6$ and zero otherwise, thus
$$P(H) = int_0.4^0.6pdp = 0.5$$
or on the other hand does it follow the standard simple uniform distribution. That is
$$P[0.4 le P le 0.6] = int_0.4^0.6 dp = 0.2$$
Then, to answer the second part, is it correctly broken down by
$$fracp^29(1-p)^21int_0.4^0.6p^29(1-p)^21dp$$
for $0.4 le p le 0.6$ and $0$ otherwise. Thus, we obtain
$$P(H|A) = int_0.4^0.6 pf(p|A)dp$$
Thank you so much for your help in advance, I really appreciate it!
probability random-variables uniform-distribution
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
2
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
2
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question states:
The probability of heads of a random coin is a random variable P, uniform in the interval $[0.4,0.6]$. Find the probability that at the next tossing of the coin that heads will show. Suppose the coin is tossed $50$ times resulting in $21$ tails and $29$ heads. What is $P(p|Observed Data)$?
My first question is what is the probability of P. It seems like there are two different ways to look at it. Is the probability such that $f(p)=5$ for $0.4 le p le 0.6$ and zero otherwise, thus
$$P(H) = int_0.4^0.6pdp = 0.5$$
or on the other hand does it follow the standard simple uniform distribution. That is
$$P[0.4 le P le 0.6] = int_0.4^0.6 dp = 0.2$$
Then, to answer the second part, is it correctly broken down by
$$fracp^29(1-p)^21int_0.4^0.6p^29(1-p)^21dp$$
for $0.4 le p le 0.6$ and $0$ otherwise. Thus, we obtain
$$P(H|A) = int_0.4^0.6 pf(p|A)dp$$
Thank you so much for your help in advance, I really appreciate it!
probability random-variables uniform-distribution
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
The question states:
The probability of heads of a random coin is a random variable P, uniform in the interval $[0.4,0.6]$. Find the probability that at the next tossing of the coin that heads will show. Suppose the coin is tossed $50$ times resulting in $21$ tails and $29$ heads. What is $P(p|Observed Data)$?
My first question is what is the probability of P. It seems like there are two different ways to look at it. Is the probability such that $f(p)=5$ for $0.4 le p le 0.6$ and zero otherwise, thus
$$P(H) = int_0.4^0.6pdp = 0.5$$
or on the other hand does it follow the standard simple uniform distribution. That is
$$P[0.4 le P le 0.6] = int_0.4^0.6 dp = 0.2$$
Then, to answer the second part, is it correctly broken down by
$$fracp^29(1-p)^21int_0.4^0.6p^29(1-p)^21dp$$
for $0.4 le p le 0.6$ and $0$ otherwise. Thus, we obtain
$$P(H|A) = int_0.4^0.6 pf(p|A)dp$$
Thank you so much for your help in advance, I really appreciate it!
probability random-variables uniform-distribution
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
edited Oct 16 '16 at 18:48
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
asked Oct 15 '16 at 1:11
Billy Thorton
2951526
2951526
2
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
2
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07
add a comment |Â
2
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
2
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07
2
2
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
2
2
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let's go step by step.
Although the result you arrive to regarding the question of what is the probability we should assign to heads before seeing the coin tosses is correct, your reasoning is not. To simplify notation I will use $theta$ instead of $P$ to denote the random variable which represents the coin bias.
First we want to find $P(H|theta is uniform in [0.4,0.6])$. But this is the same as $int_0.4^0.6 P(H|theta = x) pi_theta(x) dx$, where $pi$ is the prior density of $theta$ and it is $pi_theta(x) = fracI_[0.4,0.6](x)0.6-0.4$.
Substituting, we have that $P(H) = int_0.4^0.6 x frac10.6-0.4(x) dx=frac12$.
Now we want to do a Bayes update given the data on the coin tosses.
If we denote by $B(n, h, p)$ the mass distribution of a binomial, where $n$ is the number of coin tosses, $h$ the number of heads and $p$ the coin bias, then:
$$
f(theta = x|Data) = pi_theta(x)fractheta = x)P(Data)= pi_theta(x)fracB(50,29, x)int_0.4^0.6B(50,29, y)pi_theta(y)dy=
frac10.6-0.4frac50 choose 29 x^29 (1-x)^21int_0.4^0.650 choose 29 y^29 (1-y)^21 frac10.6-0.4dy=
frac x^29 (1-x)^21int_0.4^0.6 y^29 (1-y)^21dy
$$
Thus the result is a beta distribution with a rather hideous denominator.
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let's go step by step.
Although the result you arrive to regarding the question of what is the probability we should assign to heads before seeing the coin tosses is correct, your reasoning is not. To simplify notation I will use $theta$ instead of $P$ to denote the random variable which represents the coin bias.
First we want to find $P(H|theta is uniform in [0.4,0.6])$. But this is the same as $int_0.4^0.6 P(H|theta = x) pi_theta(x) dx$, where $pi$ is the prior density of $theta$ and it is $pi_theta(x) = fracI_[0.4,0.6](x)0.6-0.4$.
Substituting, we have that $P(H) = int_0.4^0.6 x frac10.6-0.4(x) dx=frac12$.
Now we want to do a Bayes update given the data on the coin tosses.
If we denote by $B(n, h, p)$ the mass distribution of a binomial, where $n$ is the number of coin tosses, $h$ the number of heads and $p$ the coin bias, then:
$$
f(theta = x|Data) = pi_theta(x)fractheta = x)P(Data)= pi_theta(x)fracB(50,29, x)int_0.4^0.6B(50,29, y)pi_theta(y)dy=
frac10.6-0.4frac50 choose 29 x^29 (1-x)^21int_0.4^0.650 choose 29 y^29 (1-y)^21 frac10.6-0.4dy=
frac x^29 (1-x)^21int_0.4^0.6 y^29 (1-y)^21dy
$$
Thus the result is a beta distribution with a rather hideous denominator.
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
add a comment |Â
up vote
0
down vote
Let's go step by step.
Although the result you arrive to regarding the question of what is the probability we should assign to heads before seeing the coin tosses is correct, your reasoning is not. To simplify notation I will use $theta$ instead of $P$ to denote the random variable which represents the coin bias.
First we want to find $P(H|theta is uniform in [0.4,0.6])$. But this is the same as $int_0.4^0.6 P(H|theta = x) pi_theta(x) dx$, where $pi$ is the prior density of $theta$ and it is $pi_theta(x) = fracI_[0.4,0.6](x)0.6-0.4$.
Substituting, we have that $P(H) = int_0.4^0.6 x frac10.6-0.4(x) dx=frac12$.
Now we want to do a Bayes update given the data on the coin tosses.
If we denote by $B(n, h, p)$ the mass distribution of a binomial, where $n$ is the number of coin tosses, $h$ the number of heads and $p$ the coin bias, then:
$$
f(theta = x|Data) = pi_theta(x)fractheta = x)P(Data)= pi_theta(x)fracB(50,29, x)int_0.4^0.6B(50,29, y)pi_theta(y)dy=
frac10.6-0.4frac50 choose 29 x^29 (1-x)^21int_0.4^0.650 choose 29 y^29 (1-y)^21 frac10.6-0.4dy=
frac x^29 (1-x)^21int_0.4^0.6 y^29 (1-y)^21dy
$$
Thus the result is a beta distribution with a rather hideous denominator.
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's go step by step.
Although the result you arrive to regarding the question of what is the probability we should assign to heads before seeing the coin tosses is correct, your reasoning is not. To simplify notation I will use $theta$ instead of $P$ to denote the random variable which represents the coin bias.
First we want to find $P(H|theta is uniform in [0.4,0.6])$. But this is the same as $int_0.4^0.6 P(H|theta = x) pi_theta(x) dx$, where $pi$ is the prior density of $theta$ and it is $pi_theta(x) = fracI_[0.4,0.6](x)0.6-0.4$.
Substituting, we have that $P(H) = int_0.4^0.6 x frac10.6-0.4(x) dx=frac12$.
Now we want to do a Bayes update given the data on the coin tosses.
If we denote by $B(n, h, p)$ the mass distribution of a binomial, where $n$ is the number of coin tosses, $h$ the number of heads and $p$ the coin bias, then:
$$
f(theta = x|Data) = pi_theta(x)fractheta = x)P(Data)= pi_theta(x)fracB(50,29, x)int_0.4^0.6B(50,29, y)pi_theta(y)dy=
frac10.6-0.4frac50 choose 29 x^29 (1-x)^21int_0.4^0.650 choose 29 y^29 (1-y)^21 frac10.6-0.4dy=
frac x^29 (1-x)^21int_0.4^0.6 y^29 (1-y)^21dy
$$
Thus the result is a beta distribution with a rather hideous denominator.
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
Let's go step by step.
Although the result you arrive to regarding the question of what is the probability we should assign to heads before seeing the coin tosses is correct, your reasoning is not. To simplify notation I will use $theta$ instead of $P$ to denote the random variable which represents the coin bias.
First we want to find $P(H|theta is uniform in [0.4,0.6])$. But this is the same as $int_0.4^0.6 P(H|theta = x) pi_theta(x) dx$, where $pi$ is the prior density of $theta$ and it is $pi_theta(x) = fracI_[0.4,0.6](x)0.6-0.4$.
Substituting, we have that $P(H) = int_0.4^0.6 x frac10.6-0.4(x) dx=frac12$.
Now we want to do a Bayes update given the data on the coin tosses.
If we denote by $B(n, h, p)$ the mass distribution of a binomial, where $n$ is the number of coin tosses, $h$ the number of heads and $p$ the coin bias, then:
$$
f(theta = x|Data) = pi_theta(x)fractheta = x)P(Data)= pi_theta(x)fracB(50,29, x)int_0.4^0.6B(50,29, y)pi_theta(y)dy=
frac10.6-0.4frac50 choose 29 x^29 (1-x)^21int_0.4^0.650 choose 29 y^29 (1-y)^21 frac10.6-0.4dy=
frac x^29 (1-x)^21int_0.4^0.6 y^29 (1-y)^21dy
$$
Thus the result is a beta distribution with a rather hideous denominator.
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
answered Oct 23 '16 at 13:13
Jsevillamol
3,01911124
3,01911124
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1969001%2fthe-probability-of-heads-of-a-random-coin-is-uniform-r-v-p-find-the-probabilit%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac1.2$ for $p in [.4, .6]$ (and 0 else). Note that phrases "probability of $p$" and "probability of $P$" do not make sense, since probabilities are defined on events ("Probability that $P > .5$" is something that makes sense). You likely are supposed to compute the conditional PDF of $P$: $$f_Data(p|Data=(21,29))=fracP=p]f_P(p)P[Data=(21,29)]$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
– Michael
Oct 17 '16 at 3:02
2
PS: I do not know what "H" or "A" mean in your question.
– Michael
Oct 17 '16 at 3:07