Mixing
Likelihood Function Given Maximum of data, but not actually data points
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I was just wondering how I would go about creating a likelihood function if I have a $N( theta,1)$ distribution and know $x(n)$ the maximum of n observations, but not the actual observations themselves. I understand how to create a CDF for the maximum order statistic (i.e. $F(x)^n $), but the normal distribution does not have a closed form CDF.
Am I able to find the likelihood function by taking the derivative of $F(x)^n $ (in integral form) and then evaluate the derivative at $x(n)$? Or is my reasoning flawed/or is there an easier way to do this?
Any help would be appreciated. Thank you!
maximum-likelihood order-statistics
asked Jan 25 at 18:44
David H.David H.
82
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add a comment |
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I was just wondering how I would go about creating a likelihood function if I have a $N( theta,1)$ distribution and know $x(n)$ the maximum of n observations, but not the actual observations themselves. I understand how to create a CDF for the maximum order statistic (i.e. $F(x)^n $), but the normal distribution does not have a closed form CDF.
Am I able to find the likelihood function by taking the derivative of $F(x)^n $ (in integral form) and then evaluate the derivative at $x(n)$? Or is my reasoning flawed/or is there an easier way to do this?
Any help would be appreciated. Thank you!
maximum-likelihood order-statistics
asked Jan 25 at 18:44
David H.David H.
82
$endgroup$
add a comment |
$begingroup$
I was just wondering how I would go about creating a likelihood function if I have a $N( theta,1)$ distribution and know $x(n)$ the maximum of n observations, but not the actual observations themselves. I understand how to create a CDF for the maximum order statistic (i.e. $F(x)^n $), but the normal distribution does not have a closed form CDF.
Am I able to find the likelihood function by taking the derivative of $F(x)^n $ (in integral form) and then evaluate the derivative at $x(n)$? Or is my reasoning flawed/or is there an easier way to do this?
Any help would be appreciated. Thank you!
maximum-likelihood order-statistics
asked Jan 25 at 18:44
David H.David H.
82
$endgroup$
I was just wondering how I would go about creating a likelihood function if I have a $N( theta,1)$ distribution and know $x(n)$ the maximum of n observations, but not the actual observations themselves. I understand how to create a CDF for the maximum order statistic (i.e. $F(x)^n $), but the normal distribution does not have a closed form CDF.
Am I able to find the likelihood function by taking the derivative of $F(x)^n $ (in integral form) and then evaluate the derivative at $x(n)$? Or is my reasoning flawed/or is there an easier way to do this?
Any help would be appreciated. Thank you!
maximum-likelihood order-statistics
maximum-likelihood order-statistics
asked Jan 25 at 18:44
David H.David H.
82
asked Jan 25 at 18:44
David H.David H.
82
asked Jan 25 at 18:44
David H.David H.
82
asked Jan 25 at 18:44
David H.David H.
82
asked Jan 25 at 18:44
David H.David H.
82
82
add a comment |
add a comment |
1 Answer
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We have the functions $Phi()$ and $phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,ldots,X_nsim N(theta,1)$, the distribution function of $T=maxlimits_{1le kle n}X_k$ is
$$P(Tle t)=(P(X_1le t))^n=(Phi(t-theta))^nquad,,tinmathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,tinmathbb R$$
The likelihood function given $t ,(inmathbb R)$, the observed value of $T$, is therefore
$$L(thetamid t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,thetainmathbb R$$
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
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Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have the functions $Phi()$ and $phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,ldots,X_nsim N(theta,1)$, the distribution function of $T=maxlimits_{1le kle n}X_k$ is
$$P(Tle t)=(P(X_1le t))^n=(Phi(t-theta))^nquad,,tinmathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,tinmathbb R$$
The likelihood function given $t ,(inmathbb R)$, the observed value of $T$, is therefore
$$L(thetamid t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,thetainmathbb R$$
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
$endgroup$
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
add a comment |
$begingroup$
We have the functions $Phi()$ and $phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,ldots,X_nsim N(theta,1)$, the distribution function of $T=maxlimits_{1le kle n}X_k$ is
$$P(Tle t)=(P(X_1le t))^n=(Phi(t-theta))^nquad,,tinmathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,tinmathbb R$$
The likelihood function given $t ,(inmathbb R)$, the observed value of $T$, is therefore
$$L(thetamid t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,thetainmathbb R$$
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
$endgroup$
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
add a comment |
$begingroup$
We have the functions $Phi()$ and $phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,ldots,X_nsim N(theta,1)$, the distribution function of $T=maxlimits_{1le kle n}X_k$ is
$$P(Tle t)=(P(X_1le t))^n=(Phi(t-theta))^nquad,,tinmathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,tinmathbb R$$
The likelihood function given $t ,(inmathbb R)$, the observed value of $T$, is therefore
$$L(thetamid t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,thetainmathbb R$$
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
$endgroup$
We have the functions $Phi()$ and $phi()$, standard notation for the cdf and pdf respectively of a $N(0,1)$ distribution. Or if you want to use $F$ and $f$, that also makes no difference as long as you define them. The fact that the cdf does not have a closed form does not stop us from writing down the likelihood at least.
Assuming you have i.i.d observations $X_1,ldots,X_nsim N(theta,1)$, the distribution function of $T=maxlimits_{1le kle n}X_k$ is
$$P(Tle t)=(P(X_1le t))^n=(Phi(t-theta))^nquad,,tinmathbb R$$
So the pdf of $T$ is
$$f_{T}(t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,tinmathbb R$$
The likelihood function given $t ,(inmathbb R)$, the observed value of $T$, is therefore
$$L(thetamid t)=n(Phi(t-theta))^{n-1}phi(t-theta)quad,,thetainmathbb R$$
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
edited Jan 25 at 19:19
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
answered Jan 25 at 19:08
StubbornAtomStubbornAtom
6,23821339
6,23821339
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
add a comment |
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
$begingroup$
Thank you so much! That was really clear :)
$endgroup$
– David H.
Jan 25 at 19:56
add a comment |
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