MLE of $theta$ in $U[0,theta]$ distribution where the parameter $theta$ is discrete
$begingroup$
Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the $U[0,theta]$ distribution: $$f_{theta}(x)=frac{mathbf1_{[0,theta]}(x)}{theta}$$
, where the unknown parameter $thetain{1,2,ldots}$.
What can I say about the maximum likelihood estimator (MLE) of $theta$?
Usually the parameter space is $mathbb R^{+}$ in which case the MLE is known to be $$X_{(n)}=max{X_1,X_2,ldots,X_n}$$
, but here the parameter space is restricted to the natural numbers.
In any case, the likelihood function given the sample $x_1,ldots,x_n$ is
$$L(theta)=frac{mathbf1_{[0,theta]}(x_1,ldots,x_n)}{theta^n}=frac{mathbf1_{[x_{(n)},infty)}(theta)}{theta^n}qquad,,thetainmathbb N$$
So I should check the values of $L(theta)$ for each $thetain{1,2,ldots}$ and the MLE is that value of $theta$ for which $L(theta)$ is maximized. Is this the correct strategy or can I say that MLE of $theta$ is $hattheta=lfloor X_{(n)}rfloor$? Or does the MLE exist at all? I am not sure.
Any hint would be helpful.
statistics probability-distributions uniform-distribution maximum-likelihood
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
$endgroup$
add a comment |
$begingroup$
Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the $U[0,theta]$ distribution: $$f_{theta}(x)=frac{mathbf1_{[0,theta]}(x)}{theta}$$
, where the unknown parameter $thetain{1,2,ldots}$.
What can I say about the maximum likelihood estimator (MLE) of $theta$?
Usually the parameter space is $mathbb R^{+}$ in which case the MLE is known to be $$X_{(n)}=max{X_1,X_2,ldots,X_n}$$
, but here the parameter space is restricted to the natural numbers.
In any case, the likelihood function given the sample $x_1,ldots,x_n$ is
$$L(theta)=frac{mathbf1_{[0,theta]}(x_1,ldots,x_n)}{theta^n}=frac{mathbf1_{[x_{(n)},infty)}(theta)}{theta^n}qquad,,thetainmathbb N$$
So I should check the values of $L(theta)$ for each $thetain{1,2,ldots}$ and the MLE is that value of $theta$ for which $L(theta)$ is maximized. Is this the correct strategy or can I say that MLE of $theta$ is $hattheta=lfloor X_{(n)}rfloor$? Or does the MLE exist at all? I am not sure.
Any hint would be helpful.
statistics probability-distributions uniform-distribution maximum-likelihood
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
$endgroup$
$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49
add a comment |
$begingroup$
Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the $U[0,theta]$ distribution: $$f_{theta}(x)=frac{mathbf1_{[0,theta]}(x)}{theta}$$
, where the unknown parameter $thetain{1,2,ldots}$.
What can I say about the maximum likelihood estimator (MLE) of $theta$?
Usually the parameter space is $mathbb R^{+}$ in which case the MLE is known to be $$X_{(n)}=max{X_1,X_2,ldots,X_n}$$
, but here the parameter space is restricted to the natural numbers.
In any case, the likelihood function given the sample $x_1,ldots,x_n$ is
$$L(theta)=frac{mathbf1_{[0,theta]}(x_1,ldots,x_n)}{theta^n}=frac{mathbf1_{[x_{(n)},infty)}(theta)}{theta^n}qquad,,thetainmathbb N$$
So I should check the values of $L(theta)$ for each $thetain{1,2,ldots}$ and the MLE is that value of $theta$ for which $L(theta)$ is maximized. Is this the correct strategy or can I say that MLE of $theta$ is $hattheta=lfloor X_{(n)}rfloor$? Or does the MLE exist at all? I am not sure.
Any hint would be helpful.
statistics probability-distributions uniform-distribution maximum-likelihood
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
$endgroup$
Consider i.i.d random variables $X_1,X_2,ldots,X_n$ having the $U[0,theta]$ distribution: $$f_{theta}(x)=frac{mathbf1_{[0,theta]}(x)}{theta}$$
, where the unknown parameter $thetain{1,2,ldots}$.
What can I say about the maximum likelihood estimator (MLE) of $theta$?
Usually the parameter space is $mathbb R^{+}$ in which case the MLE is known to be $$X_{(n)}=max{X_1,X_2,ldots,X_n}$$
, but here the parameter space is restricted to the natural numbers.
In any case, the likelihood function given the sample $x_1,ldots,x_n$ is
$$L(theta)=frac{mathbf1_{[0,theta]}(x_1,ldots,x_n)}{theta^n}=frac{mathbf1_{[x_{(n)},infty)}(theta)}{theta^n}qquad,,thetainmathbb N$$
So I should check the values of $L(theta)$ for each $thetain{1,2,ldots}$ and the MLE is that value of $theta$ for which $L(theta)$ is maximized. Is this the correct strategy or can I say that MLE of $theta$ is $hattheta=lfloor X_{(n)}rfloor$? Or does the MLE exist at all? I am not sure.
Any hint would be helpful.
statistics probability-distributions uniform-distribution maximum-likelihood
statistics probability-distributions uniform-distribution maximum-likelihood
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
asked Dec 14 '18 at 17:00
StubbornAtomStubbornAtom
6,06811239
6,06811239
$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49
add a comment |
$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49
$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49
add a comment |
1 Answer
1
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$begingroup$
Given a sample $xequiv {x_i}_{i=1}^n$, the likelihood is
$$
L(thetamid x)=theta^{-n}1{thetage M(x),m(x)ge 0},
$$
where $M(x):=lceilmax_{1le ile n}x_irceil$ and $m(x):=min_{1le ile n}x_i$.
The indicator suggests that $hat{theta}_n(x)ge M(x)$ ($because$ $L=0$ otherwise). However, taking values larger than $M(x)$ decreases $L$ because of the first term (assuming that $m(x)>0$). Thus, $hat{theta}_n(x)= M(x)$.
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
$endgroup$
add a comment |
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$begingroup$
Given a sample $xequiv {x_i}_{i=1}^n$, the likelihood is
$$
L(thetamid x)=theta^{-n}1{thetage M(x),m(x)ge 0},
$$
where $M(x):=lceilmax_{1le ile n}x_irceil$ and $m(x):=min_{1le ile n}x_i$.
The indicator suggests that $hat{theta}_n(x)ge M(x)$ ($because$ $L=0$ otherwise). However, taking values larger than $M(x)$ decreases $L$ because of the first term (assuming that $m(x)>0$). Thus, $hat{theta}_n(x)= M(x)$.
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
$endgroup$
add a comment |
$begingroup$
Given a sample $xequiv {x_i}_{i=1}^n$, the likelihood is
$$
L(thetamid x)=theta^{-n}1{thetage M(x),m(x)ge 0},
$$
where $M(x):=lceilmax_{1le ile n}x_irceil$ and $m(x):=min_{1le ile n}x_i$.
The indicator suggests that $hat{theta}_n(x)ge M(x)$ ($because$ $L=0$ otherwise). However, taking values larger than $M(x)$ decreases $L$ because of the first term (assuming that $m(x)>0$). Thus, $hat{theta}_n(x)= M(x)$.
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
$endgroup$
add a comment |
$begingroup$
Given a sample $xequiv {x_i}_{i=1}^n$, the likelihood is
$$
L(thetamid x)=theta^{-n}1{thetage M(x),m(x)ge 0},
$$
where $M(x):=lceilmax_{1le ile n}x_irceil$ and $m(x):=min_{1le ile n}x_i$.
The indicator suggests that $hat{theta}_n(x)ge M(x)$ ($because$ $L=0$ otherwise). However, taking values larger than $M(x)$ decreases $L$ because of the first term (assuming that $m(x)>0$). Thus, $hat{theta}_n(x)= M(x)$.
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
$endgroup$
Given a sample $xequiv {x_i}_{i=1}^n$, the likelihood is
$$
L(thetamid x)=theta^{-n}1{thetage M(x),m(x)ge 0},
$$
where $M(x):=lceilmax_{1le ile n}x_irceil$ and $m(x):=min_{1le ile n}x_i$.
The indicator suggests that $hat{theta}_n(x)ge M(x)$ ($because$ $L=0$ otherwise). However, taking values larger than $M(x)$ decreases $L$ because of the first term (assuming that $m(x)>0$). Thus, $hat{theta}_n(x)= M(x)$.
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
edited Feb 21 at 17:21
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
answered Dec 14 '18 at 18:21
d.k.o.d.k.o.
10k629
10k629
add a comment |
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$begingroup$
$theta$ must be at least as big as the largest $X_i$ so surely it should equal the next largest integer.
$endgroup$
– user121049
Dec 14 '18 at 17:45
$begingroup$
@user121049 Yes, I should have used 'ceiling' instead of 'floor'.
$endgroup$
– StubbornAtom
Dec 14 '18 at 17:49