The probability of heads of a random coin is uniform r.v. P. Find the probability that heads will show?












0












$begingroup$


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
$$P(H) = int_{0.4}^{0.6}pdp = 0.5$$
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
$$frac{p^{29}(1-p)^{21}}{int_{0.4}^{0.6}p^{29}(1-p)^{21}dp}$$
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







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  • 2




    $begingroup$
    Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:02








  • 2




    $begingroup$
    PS: I do not know what "H" or "A" mean in your question.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:07
















0












$begingroup$


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
$$P(H) = int_{0.4}^{0.6}pdp = 0.5$$
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
$$frac{p^{29}(1-p)^{21}}{int_{0.4}^{0.6}p^{29}(1-p)^{21}dp}$$
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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:02








  • 2




    $begingroup$
    PS: I do not know what "H" or "A" mean in your question.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:07














0












0








0





$begingroup$


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
$$P(H) = int_{0.4}^{0.6}pdp = 0.5$$
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
$$frac{p^{29}(1-p)^{21}}{int_{0.4}^{0.6}p^{29}(1-p)^{21}dp}$$
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







share|cite|improve this question











$endgroup$




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
$$P(H) = int_{0.4}^{0.6}pdp = 0.5$$
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
$$frac{p^{29}(1-p)^{21}}{int_{0.4}^{0.6}p^{29}(1-p)^{21}dp}$$
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




probability random-variables uniform-distribution






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edited Oct 16 '16 at 18:48







Billy Thorton

















asked Oct 15 '16 at 1:11









Billy ThortonBilly Thorton

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  • 2




    $begingroup$
    Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:02








  • 2




    $begingroup$
    PS: I do not know what "H" or "A" mean in your question.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:07














  • 2




    $begingroup$
    Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:02








  • 2




    $begingroup$
    PS: I do not know what "H" or "A" mean in your question.
    $endgroup$
    – Michael
    Oct 17 '16 at 3:07








2




2




$begingroup$
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
$endgroup$
– Michael
Oct 17 '16 at 3:02






$begingroup$
Your random variable $P$ has a uniform distribution over $[.4, .6]$, and so $P[.4 leq Pleq .6]=1$. The PDF of $P$ is $f_P(p) = frac{1}{.2}$ 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 $P$: $$f_{P|Data}(p|Data=(21,29))=frac{P[Data=(21,29)|P=p]f_P(p)}{P[Data=(21,29)]}$$ and compute $P[Data=(21,29)]$ by conditioning on $P=p$ via the law of total probability.
$endgroup$
– Michael
Oct 17 '16 at 3:02






2




2




$begingroup$
PS: I do not know what "H" or "A" mean in your question.
$endgroup$
– Michael
Oct 17 '16 at 3:07




$begingroup$
PS: I do not know what "H" or "A" mean in your question.
$endgroup$
– Michael
Oct 17 '16 at 3:07










1 Answer
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0












$begingroup$

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 $pi_theta(x) = frac{I_{[0.4,0.6]}(x)}{0.6-0.4}$.



Substituting, we have that $P(H) = int_{0.4}^{0.6} x frac{1}{0.6-0.4}(x) dx=frac{1}{2}$.




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:
$$
f(theta = x|Data) = pi_{theta}(x)frac{P(Data|theta = x)}{P(Data)}= pi_{theta}(x)frac{B(50,29, x)}{int_{0.4}^{0.6}B(50,29, y)pi_theta(y)dy}=
frac{1}{0.6-0.4}frac{{50 choose 29} x^{29} (1-x)^{21}}{int_{0.4}^{0.6}{50 choose 29} y^{29} (1-y)^{21} frac{1}{0.6-0.4}dy}=
frac{ x^{29} (1-x)^{21}}{int_{0.4}^{0.6} y^{29} (1-y)^{21}dy}
$$
Thus the result is a beta distribution with a rather hideous denominator.






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    1 Answer
    1






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    active

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    active

    oldest

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    0












    $begingroup$

    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 $pi_theta(x) = frac{I_{[0.4,0.6]}(x)}{0.6-0.4}$.



    Substituting, we have that $P(H) = int_{0.4}^{0.6} x frac{1}{0.6-0.4}(x) dx=frac{1}{2}$.




    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:
    $$
    f(theta = x|Data) = pi_{theta}(x)frac{P(Data|theta = x)}{P(Data)}= pi_{theta}(x)frac{B(50,29, x)}{int_{0.4}^{0.6}B(50,29, y)pi_theta(y)dy}=
    frac{1}{0.6-0.4}frac{{50 choose 29} x^{29} (1-x)^{21}}{int_{0.4}^{0.6}{50 choose 29} y^{29} (1-y)^{21} frac{1}{0.6-0.4}dy}=
    frac{ x^{29} (1-x)^{21}}{int_{0.4}^{0.6} y^{29} (1-y)^{21}dy}
    $$
    Thus the result is a beta distribution with a rather hideous denominator.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $pi_theta(x) = frac{I_{[0.4,0.6]}(x)}{0.6-0.4}$.



      Substituting, we have that $P(H) = int_{0.4}^{0.6} x frac{1}{0.6-0.4}(x) dx=frac{1}{2}$.




      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:
      $$
      f(theta = x|Data) = pi_{theta}(x)frac{P(Data|theta = x)}{P(Data)}= pi_{theta}(x)frac{B(50,29, x)}{int_{0.4}^{0.6}B(50,29, y)pi_theta(y)dy}=
      frac{1}{0.6-0.4}frac{{50 choose 29} x^{29} (1-x)^{21}}{int_{0.4}^{0.6}{50 choose 29} y^{29} (1-y)^{21} frac{1}{0.6-0.4}dy}=
      frac{ x^{29} (1-x)^{21}}{int_{0.4}^{0.6} y^{29} (1-y)^{21}dy}
      $$
      Thus the result is a beta distribution with a rather hideous denominator.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $pi_theta(x) = frac{I_{[0.4,0.6]}(x)}{0.6-0.4}$.



        Substituting, we have that $P(H) = int_{0.4}^{0.6} x frac{1}{0.6-0.4}(x) dx=frac{1}{2}$.




        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:
        $$
        f(theta = x|Data) = pi_{theta}(x)frac{P(Data|theta = x)}{P(Data)}= pi_{theta}(x)frac{B(50,29, x)}{int_{0.4}^{0.6}B(50,29, y)pi_theta(y)dy}=
        frac{1}{0.6-0.4}frac{{50 choose 29} x^{29} (1-x)^{21}}{int_{0.4}^{0.6}{50 choose 29} y^{29} (1-y)^{21} frac{1}{0.6-0.4}dy}=
        frac{ x^{29} (1-x)^{21}}{int_{0.4}^{0.6} y^{29} (1-y)^{21}dy}
        $$
        Thus the result is a beta distribution with a rather hideous denominator.






        share|cite|improve this answer









        $endgroup$



        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 $pi_theta(x) = frac{I_{[0.4,0.6]}(x)}{0.6-0.4}$.



        Substituting, we have that $P(H) = int_{0.4}^{0.6} x frac{1}{0.6-0.4}(x) dx=frac{1}{2}$.




        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:
        $$
        f(theta = x|Data) = pi_{theta}(x)frac{P(Data|theta = x)}{P(Data)}= pi_{theta}(x)frac{B(50,29, x)}{int_{0.4}^{0.6}B(50,29, y)pi_theta(y)dy}=
        frac{1}{0.6-0.4}frac{{50 choose 29} x^{29} (1-x)^{21}}{int_{0.4}^{0.6}{50 choose 29} y^{29} (1-y)^{21} frac{1}{0.6-0.4}dy}=
        frac{ x^{29} (1-x)^{21}}{int_{0.4}^{0.6} y^{29} (1-y)^{21}dy}
        $$
        Thus the result is a beta distribution with a rather hideous denominator.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 23 '16 at 13:13









        JsevillamolJsevillamol

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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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