Expected Value of the Probability












1












$begingroup$


A paper I was reading they gave the following result without explanation:



Let $sinleft{0,1right}^n$ where the probability of 1 occurring is p. Let k be the number of 1s occurring in s.
Then $mathbf{E}[p|k]=frac{k+1}{n+2}$ .



Intuitively, I would assume $f(p)=p^k*(1-p)^{n-k}$ to be the probability density function. But then I get $mathbf{E}[p|k]=int_{0}^{1}x^{k+1}*(1-x)^{n-k}$ which yields a different result.



I'm thankful for any tips! On a side note, I've only done first year probability, hence I'm not very capable.






probability statistics expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
    $endgroup$
    – Matthew Towers
    Jan 2 at 12:06










  • $begingroup$
    the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
    $endgroup$
    – Martín Vacas Vignolo
    Jan 2 at 12:30


















1












$begingroup$


A paper I was reading they gave the following result without explanation:



Let $sinleft{0,1right}^n$ where the probability of 1 occurring is p. Let k be the number of 1s occurring in s.
Then $mathbf{E}[p|k]=frac{k+1}{n+2}$ .



Intuitively, I would assume $f(p)=p^k*(1-p)^{n-k}$ to be the probability density function. But then I get $mathbf{E}[p|k]=int_{0}^{1}x^{k+1}*(1-x)^{n-k}$ which yields a different result.



I'm thankful for any tips! On a side note, I've only done first year probability, hence I'm not very capable.






probability statistics expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
    $endgroup$
    – Matthew Towers
    Jan 2 at 12:06










  • $begingroup$
    the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
    $endgroup$
    – Martín Vacas Vignolo
    Jan 2 at 12:30
















1












1








1





$begingroup$


A paper I was reading they gave the following result without explanation:



Let $sinleft{0,1right}^n$ where the probability of 1 occurring is p. Let k be the number of 1s occurring in s.
Then $mathbf{E}[p|k]=frac{k+1}{n+2}$ .



Intuitively, I would assume $f(p)=p^k*(1-p)^{n-k}$ to be the probability density function. But then I get $mathbf{E}[p|k]=int_{0}^{1}x^{k+1}*(1-x)^{n-k}$ which yields a different result.



I'm thankful for any tips! On a side note, I've only done first year probability, hence I'm not very capable.






probability statistics expected-value







share|cite|improve this question









$endgroup$




A paper I was reading they gave the following result without explanation:



Let $sinleft{0,1right}^n$ where the probability of 1 occurring is p. Let k be the number of 1s occurring in s.
Then $mathbf{E}[p|k]=frac{k+1}{n+2}$ .



Intuitively, I would assume $f(p)=p^k*(1-p)^{n-k}$ to be the probability density function. But then I get $mathbf{E}[p|k]=int_{0}^{1}x^{k+1}*(1-x)^{n-k}$ which yields a different result.



I'm thankful for any tips! On a side note, I've only done first year probability, hence I'm not very capable.






probability statistics expected-value




probability statistics expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 11:07









AlisonAlison

164




164












  • $begingroup$
    That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
    $endgroup$
    – Matthew Towers
    Jan 2 at 12:06










  • $begingroup$
    the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
    $endgroup$
    – Martín Vacas Vignolo
    Jan 2 at 12:30




















  • $begingroup$
    That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
    $endgroup$
    – Matthew Towers
    Jan 2 at 12:06










  • $begingroup$
    the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
    $endgroup$
    – Martín Vacas Vignolo
    Jan 2 at 12:30


















$begingroup$
That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
$endgroup$
– Matthew Towers
Jan 2 at 12:06




$begingroup$
That's the expected value of the Bayesian estimator for p given a uniform prior - have a look at math.stackexchange.com/questions/2643324/…
$endgroup$
– Matthew Towers
Jan 2 at 12:06












$begingroup$
the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
$endgroup$
– Martín Vacas Vignolo
Jan 2 at 12:30






$begingroup$
the problem is in the order of 0 and 1. For example, for $k = 1$ you are counting $p(1-p)^{n-1}$, but in reality there are $n$ possible s's with this probability (hint: binomial)
$endgroup$
– Martín Vacas Vignolo
Jan 2 at 12:30












1 Answer
1






active

oldest

votes


















1












$begingroup$

Your $f(p)$ is not a probability density function but is proportional to the likelihood. So if you have a uniform prior on $[0,1]$ then $f(p)$ is also proportional to the posterior; the scaling factor is essentially a Beta function so the probability density function would be $dfrac{p^k(1-p)^{n-k}}{B(k+1,n-k+1)}$.



The mean of this posterior distribution would then be $$dfrac{int_0^1 p cdot p^k(1-p)^{n-k} , dp}{int_0^1 quad p^k(1-p)^{n-k} , dp} = dfrac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=dfrac{k+1}{n+2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
    $endgroup$
    – Alison
    Jan 3 at 12:30










  • $begingroup$
    @Alison: Yes - that is an essential feature of a probability distribution
    $endgroup$
    – Henry
    Jan 3 at 14:09











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1 Answer
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$begingroup$

Your $f(p)$ is not a probability density function but is proportional to the likelihood. So if you have a uniform prior on $[0,1]$ then $f(p)$ is also proportional to the posterior; the scaling factor is essentially a Beta function so the probability density function would be $dfrac{p^k(1-p)^{n-k}}{B(k+1,n-k+1)}$.



The mean of this posterior distribution would then be $$dfrac{int_0^1 p cdot p^k(1-p)^{n-k} , dp}{int_0^1 quad p^k(1-p)^{n-k} , dp} = dfrac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=dfrac{k+1}{n+2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
    $endgroup$
    – Alison
    Jan 3 at 12:30










  • $begingroup$
    @Alison: Yes - that is an essential feature of a probability distribution
    $endgroup$
    – Henry
    Jan 3 at 14:09
















1












$begingroup$

Your $f(p)$ is not a probability density function but is proportional to the likelihood. So if you have a uniform prior on $[0,1]$ then $f(p)$ is also proportional to the posterior; the scaling factor is essentially a Beta function so the probability density function would be $dfrac{p^k(1-p)^{n-k}}{B(k+1,n-k+1)}$.



The mean of this posterior distribution would then be $$dfrac{int_0^1 p cdot p^k(1-p)^{n-k} , dp}{int_0^1 quad p^k(1-p)^{n-k} , dp} = dfrac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=dfrac{k+1}{n+2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
    $endgroup$
    – Alison
    Jan 3 at 12:30










  • $begingroup$
    @Alison: Yes - that is an essential feature of a probability distribution
    $endgroup$
    – Henry
    Jan 3 at 14:09














1












1








1





$begingroup$

Your $f(p)$ is not a probability density function but is proportional to the likelihood. So if you have a uniform prior on $[0,1]$ then $f(p)$ is also proportional to the posterior; the scaling factor is essentially a Beta function so the probability density function would be $dfrac{p^k(1-p)^{n-k}}{B(k+1,n-k+1)}$.



The mean of this posterior distribution would then be $$dfrac{int_0^1 p cdot p^k(1-p)^{n-k} , dp}{int_0^1 quad p^k(1-p)^{n-k} , dp} = dfrac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=dfrac{k+1}{n+2}$$






share|cite|improve this answer









$endgroup$



Your $f(p)$ is not a probability density function but is proportional to the likelihood. So if you have a uniform prior on $[0,1]$ then $f(p)$ is also proportional to the posterior; the scaling factor is essentially a Beta function so the probability density function would be $dfrac{p^k(1-p)^{n-k}}{B(k+1,n-k+1)}$.



The mean of this posterior distribution would then be $$dfrac{int_0^1 p cdot p^k(1-p)^{n-k} , dp}{int_0^1 quad p^k(1-p)^{n-k} , dp} = dfrac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=dfrac{k+1}{n+2}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 14:34









HenryHenry

101k481168




101k481168












  • $begingroup$
    Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
    $endgroup$
    – Alison
    Jan 3 at 12:30










  • $begingroup$
    @Alison: Yes - that is an essential feature of a probability distribution
    $endgroup$
    – Henry
    Jan 3 at 14:09


















  • $begingroup$
    Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
    $endgroup$
    – Alison
    Jan 3 at 12:30










  • $begingroup$
    @Alison: Yes - that is an essential feature of a probability distribution
    $endgroup$
    – Henry
    Jan 3 at 14:09
















$begingroup$
Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
$endgroup$
– Alison
Jan 3 at 12:30




$begingroup$
Thank you! The intuition is that p has a probability of 1 of being in the interval [0,1] and hence we scale $p^k*(1-p)^{n-k}$ to obtain the probability density function with the integral over [0,1] equalling 1, correct?
$endgroup$
– Alison
Jan 3 at 12:30












$begingroup$
@Alison: Yes - that is an essential feature of a probability distribution
$endgroup$
– Henry
Jan 3 at 14:09




$begingroup$
@Alison: Yes - that is an essential feature of a probability distribution
$endgroup$
– Henry
Jan 3 at 14:09


















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