Estimate how many tracks the city has.(Almost done, but can not find whether estimator is biased or not)
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I try to estimate the number of tracks in the city by observing their serial numbers. Assume that the serial numbers are drawn from a uniform probability density ranging from 0 to an unknown parameter $theta$, as which I take the number of trucks in the city. I use maximum likelihood method to estimate the unknown parameter.
I observe L trucks, take their serial numbers(s1, s2, ...,sL) down. Assume the observation results are independent, the maximum likelihood estimation of $theta$ is $hattheta$ = maxs1, s2,..., sL.
To find estimator baised or not, I take expectation of $hattheta$.
$$E(hattheta)=sum_i=0^theta ifrac(i+1)^L-i^L)(theta+1)^L$$
where $frac(i+1)^L-i^L)(theta+1)^L$ is the ditribution of estimator $hattheta$.
The problem is I can not find a closed form formula for the expectation. Anyone can hep?
probability-distributions expectation maximum-likelihood estimation-theory
asked Aug 16 at 3:39
Chuan Huang
51
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up vote
0
down vote
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I try to estimate the number of tracks in the city by observing their serial numbers. Assume that the serial numbers are drawn from a uniform probability density ranging from 0 to an unknown parameter $theta$, as which I take the number of trucks in the city. I use maximum likelihood method to estimate the unknown parameter.
I observe L trucks, take their serial numbers(s1, s2, ...,sL) down. Assume the observation results are independent, the maximum likelihood estimation of $theta$ is $hattheta$ = maxs1, s2,..., sL.
To find estimator baised or not, I take expectation of $hattheta$.
$$E(hattheta)=sum_i=0^theta ifrac(i+1)^L-i^L)(theta+1)^L$$
where $frac(i+1)^L-i^L)(theta+1)^L$ is the ditribution of estimator $hattheta$.
The problem is I can not find a closed form formula for the expectation. Anyone can hep?
probability-distributions expectation maximum-likelihood estimation-theory
asked Aug 16 at 3:39
Chuan Huang
51
I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55
add a comment |Â
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I try to estimate the number of tracks in the city by observing their serial numbers. Assume that the serial numbers are drawn from a uniform probability density ranging from 0 to an unknown parameter $theta$, as which I take the number of trucks in the city. I use maximum likelihood method to estimate the unknown parameter.
I observe L trucks, take their serial numbers(s1, s2, ...,sL) down. Assume the observation results are independent, the maximum likelihood estimation of $theta$ is $hattheta$ = maxs1, s2,..., sL.
To find estimator baised or not, I take expectation of $hattheta$.
$$E(hattheta)=sum_i=0^theta ifrac(i+1)^L-i^L)(theta+1)^L$$
where $frac(i+1)^L-i^L)(theta+1)^L$ is the ditribution of estimator $hattheta$.
The problem is I can not find a closed form formula for the expectation. Anyone can hep?
probability-distributions expectation maximum-likelihood estimation-theory
asked Aug 16 at 3:39
Chuan Huang
51
I try to estimate the number of tracks in the city by observing their serial numbers. Assume that the serial numbers are drawn from a uniform probability density ranging from 0 to an unknown parameter $theta$, as which I take the number of trucks in the city. I use maximum likelihood method to estimate the unknown parameter.
I observe L trucks, take their serial numbers(s1, s2, ...,sL) down. Assume the observation results are independent, the maximum likelihood estimation of $theta$ is $hattheta$ = maxs1, s2,..., sL.
To find estimator baised or not, I take expectation of $hattheta$.
$$E(hattheta)=sum_i=0^theta ifrac(i+1)^L-i^L)(theta+1)^L$$
where $frac(i+1)^L-i^L)(theta+1)^L$ is the ditribution of estimator $hattheta$.
The problem is I can not find a closed form formula for the expectation. Anyone can hep?
probability-distributions expectation maximum-likelihood estimation-theory
asked Aug 16 at 3:39
Chuan Huang
51
asked Aug 16 at 3:39
Chuan Huang
51
asked Aug 16 at 3:39
Chuan Huang
51
asked Aug 16 at 3:39
Chuan Huang
51
51
I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55
add a comment |Â
I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55
I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55
add a comment |Â
1 Answer
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For independent and identically distributed observations $boldsymbol s = (s_1, ldots, s_n)$ from a discrete uniform distribution $S$ on $1, 2, ldots, theta$, the estimator $$hat theta = max_i s_i$$ has the probability mass function
$$beginalign* Pr[hat theta = x]
&= Pr[hat theta le x] - Pr[hat theta le x-1] \
&= prod_i=1^n Pr[S_i le x] - prod_i=1^n Pr[S_i le x-1] \
&= left(fracxthetaright)^n - left(fracx-1thetaright)^n, quad x in 1, 2, ldots, theta, quad n in mathbb Z^+.
endalign*$$
The expectation is then
$$beginalign* operatornameE[hattheta]
&= sum_i=1^theta x Pr[hat theta = x] \
&= theta^-n sum_x=1^theta x(x^n - (x-1)^n) \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 (x+1)x^n \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 x^n+1 + x^n \
&= theta^-n left( theta^n+1 - 0^n+1 - sum_x=1^theta-1 x^n right) \
&= theta - fracH_theta-1,-ntheta^n \
endalign*$$
where $H_m,n = sum_x=1^m frac1x^n$ is a generalized harmonic number. It is trivial to see that $operatornameE[hat theta] < theta$, since $$fracH_theta-1,-ntheta^n = sum_x=1^theta-1 left(fracxthetaright)^n$$ is clearly a finite sum of positive rationals.
It is worth noting that the above calculation is completely unnecessary to establish that $hat theta$ is necessarily biased, since $$Pr[hat theta > theta] = 0,$$ yet $$Pr[hat theta < theta] > 0.$$
answered Aug 16 at 5:08
heropup
59.9k65895
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
accepted
For independent and identically distributed observations $boldsymbol s = (s_1, ldots, s_n)$ from a discrete uniform distribution $S$ on $1, 2, ldots, theta$, the estimator $$hat theta = max_i s_i$$ has the probability mass function
$$beginalign* Pr[hat theta = x]
&= Pr[hat theta le x] - Pr[hat theta le x-1] \
&= prod_i=1^n Pr[S_i le x] - prod_i=1^n Pr[S_i le x-1] \
&= left(fracxthetaright)^n - left(fracx-1thetaright)^n, quad x in 1, 2, ldots, theta, quad n in mathbb Z^+.
endalign*$$
The expectation is then
$$beginalign* operatornameE[hattheta]
&= sum_i=1^theta x Pr[hat theta = x] \
&= theta^-n sum_x=1^theta x(x^n - (x-1)^n) \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 (x+1)x^n \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 x^n+1 + x^n \
&= theta^-n left( theta^n+1 - 0^n+1 - sum_x=1^theta-1 x^n right) \
&= theta - fracH_theta-1,-ntheta^n \
endalign*$$
where $H_m,n = sum_x=1^m frac1x^n$ is a generalized harmonic number. It is trivial to see that $operatornameE[hat theta] < theta$, since $$fracH_theta-1,-ntheta^n = sum_x=1^theta-1 left(fracxthetaright)^n$$ is clearly a finite sum of positive rationals.
It is worth noting that the above calculation is completely unnecessary to establish that $hat theta$ is necessarily biased, since $$Pr[hat theta > theta] = 0,$$ yet $$Pr[hat theta < theta] > 0.$$
answered Aug 16 at 5:08
heropup
59.9k65895
add a comment |Â
up vote
0
down vote
accepted
For independent and identically distributed observations $boldsymbol s = (s_1, ldots, s_n)$ from a discrete uniform distribution $S$ on $1, 2, ldots, theta$, the estimator $$hat theta = max_i s_i$$ has the probability mass function
$$beginalign* Pr[hat theta = x]
&= Pr[hat theta le x] - Pr[hat theta le x-1] \
&= prod_i=1^n Pr[S_i le x] - prod_i=1^n Pr[S_i le x-1] \
&= left(fracxthetaright)^n - left(fracx-1thetaright)^n, quad x in 1, 2, ldots, theta, quad n in mathbb Z^+.
endalign*$$
The expectation is then
$$beginalign* operatornameE[hattheta]
&= sum_i=1^theta x Pr[hat theta = x] \
&= theta^-n sum_x=1^theta x(x^n - (x-1)^n) \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 (x+1)x^n \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 x^n+1 + x^n \
&= theta^-n left( theta^n+1 - 0^n+1 - sum_x=1^theta-1 x^n right) \
&= theta - fracH_theta-1,-ntheta^n \
endalign*$$
where $H_m,n = sum_x=1^m frac1x^n$ is a generalized harmonic number. It is trivial to see that $operatornameE[hat theta] < theta$, since $$fracH_theta-1,-ntheta^n = sum_x=1^theta-1 left(fracxthetaright)^n$$ is clearly a finite sum of positive rationals.
It is worth noting that the above calculation is completely unnecessary to establish that $hat theta$ is necessarily biased, since $$Pr[hat theta > theta] = 0,$$ yet $$Pr[hat theta < theta] > 0.$$
answered Aug 16 at 5:08
heropup
59.9k65895
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For independent and identically distributed observations $boldsymbol s = (s_1, ldots, s_n)$ from a discrete uniform distribution $S$ on $1, 2, ldots, theta$, the estimator $$hat theta = max_i s_i$$ has the probability mass function
$$beginalign* Pr[hat theta = x]
&= Pr[hat theta le x] - Pr[hat theta le x-1] \
&= prod_i=1^n Pr[S_i le x] - prod_i=1^n Pr[S_i le x-1] \
&= left(fracxthetaright)^n - left(fracx-1thetaright)^n, quad x in 1, 2, ldots, theta, quad n in mathbb Z^+.
endalign*$$
The expectation is then
$$beginalign* operatornameE[hattheta]
&= sum_i=1^theta x Pr[hat theta = x] \
&= theta^-n sum_x=1^theta x(x^n - (x-1)^n) \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 (x+1)x^n \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 x^n+1 + x^n \
&= theta^-n left( theta^n+1 - 0^n+1 - sum_x=1^theta-1 x^n right) \
&= theta - fracH_theta-1,-ntheta^n \
endalign*$$
where $H_m,n = sum_x=1^m frac1x^n$ is a generalized harmonic number. It is trivial to see that $operatornameE[hat theta] < theta$, since $$fracH_theta-1,-ntheta^n = sum_x=1^theta-1 left(fracxthetaright)^n$$ is clearly a finite sum of positive rationals.
It is worth noting that the above calculation is completely unnecessary to establish that $hat theta$ is necessarily biased, since $$Pr[hat theta > theta] = 0,$$ yet $$Pr[hat theta < theta] > 0.$$
answered Aug 16 at 5:08
heropup
59.9k65895
For independent and identically distributed observations $boldsymbol s = (s_1, ldots, s_n)$ from a discrete uniform distribution $S$ on $1, 2, ldots, theta$, the estimator $$hat theta = max_i s_i$$ has the probability mass function
$$beginalign* Pr[hat theta = x]
&= Pr[hat theta le x] - Pr[hat theta le x-1] \
&= prod_i=1^n Pr[S_i le x] - prod_i=1^n Pr[S_i le x-1] \
&= left(fracxthetaright)^n - left(fracx-1thetaright)^n, quad x in 1, 2, ldots, theta, quad n in mathbb Z^+.
endalign*$$
The expectation is then
$$beginalign* operatornameE[hattheta]
&= sum_i=1^theta x Pr[hat theta = x] \
&= theta^-n sum_x=1^theta x(x^n - (x-1)^n) \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 (x+1)x^n \
&= theta^-n sum_x=1^theta x^n+1 - sum_x=0^theta-1 x^n+1 + x^n \
&= theta^-n left( theta^n+1 - 0^n+1 - sum_x=1^theta-1 x^n right) \
&= theta - fracH_theta-1,-ntheta^n \
endalign*$$
where $H_m,n = sum_x=1^m frac1x^n$ is a generalized harmonic number. It is trivial to see that $operatornameE[hat theta] < theta$, since $$fracH_theta-1,-ntheta^n = sum_x=1^theta-1 left(fracxthetaright)^n$$ is clearly a finite sum of positive rationals.
It is worth noting that the above calculation is completely unnecessary to establish that $hat theta$ is necessarily biased, since $$Pr[hat theta > theta] = 0,$$ yet $$Pr[hat theta < theta] > 0.$$
answered Aug 16 at 5:08
heropup
59.9k65895
edited Aug 16 at 5:17
answered Aug 16 at 5:08
heropup
59.9k65895
answered Aug 16 at 5:08
heropup
59.9k65895
answered Aug 16 at 5:08
heropup
59.9k65895
59.9k65895
add a comment |Â
add a comment |Â
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I seem to recall that your estimate should be $frac L+1L$ times the highest serial number observed.
– Ross Millikan
Aug 16 at 3:50
@RossMillikan I don't understand. Could you show me why is that?
– Chuan Huang
Aug 16 at 6:35
It is because if you pick L items independently from a uniform distribution the expected value of the largest is $frac LL+1$. I don't have a demonstration of that, but it is a standard result.
– Ross Millikan
Aug 16 at 14:29
You might be interested in the "German tank problem". en.wikipedia.org/wiki/German_tank_problem
– awkward
Aug 16 at 18:55