Fdtxjr

ssTTsSTtRrriinInnnnNNNIiinngg

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How to find the MLE of these parameters given distribution?

Minimally Sufficient Statistic for Bivariate DistributionAssuming $sigma$ is known, find a method of moments estimator of $mu$. (Location-scale family of exponential distribution)finding the maximum likelihood estimator of location scale family of exponential distributionsComputation of a likelihood for a discrete variableDistribution of minimum of 2 INID random variables when one is transformed.Find joint density function of X and X+Y (exponential distribution)Maximum a Posteriori estimation of the parameter of a exponential distribution with a gaussian priorfind joint distribution $min (X,Y)$ and $W$, where W is a discrete distributionWhy are cross expectations zero in MLE?how fitting a Gaussian approximation to the likelihood curve at maximum?

1

0

$begingroup$

Let $X$ and $Y$ be independent exponential random variables, with

$$f(xmidlambda)=frac1lambdaexpleft(-fracxlambdaright),,x>0,, qquad f(ymidmu)=frac1muexpleft(-fracymuright),,y>0$$

We observe $Z$ and $W$ with $Z=min(X,Y)$, and $W=begincases 1 &,textif Z=X\ 0 &,textif Z=Y endcases$

I have obtained the joint distribution of $Z$ and $W$, i.e.,
$$P(Z leq z, W=0)=fraclambdamu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

$$P(Z leq z, W=1)=fracmumu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

Now assume that $(Z_i,W_i),i=1,cdots,n$, are $n$ i.i.d observations. Find the MLEs of $lambda$ and $mu$.

(This is the exercise 7.14 of the book Statistical Inference 2nd edition, but no solution given)

statistics maximum-likelihood exponential-distribution censoring

share|cite|improve this question

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

$endgroup$

add a comment | 

1

0

$begingroup$

Let $X$ and $Y$ be independent exponential random variables, with

$$f(xmidlambda)=frac1lambdaexpleft(-fracxlambdaright),,x>0,, qquad f(ymidmu)=frac1muexpleft(-fracymuright),,y>0$$

We observe $Z$ and $W$ with $Z=min(X,Y)$, and $W=begincases 1 &,textif Z=X\ 0 &,textif Z=Y endcases$

I have obtained the joint distribution of $Z$ and $W$, i.e.,
$$P(Z leq z, W=0)=fraclambdamu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

$$P(Z leq z, W=1)=fracmumu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

Now assume that $(Z_i,W_i),i=1,cdots,n$, are $n$ i.i.d observations. Find the MLEs of $lambda$ and $mu$.

(This is the exercise 7.14 of the book Statistical Inference 2nd edition, but no solution given)

statistics maximum-likelihood exponential-distribution censoring

share|cite|improve this question

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

$endgroup$

add a comment | 

$begingroup$

Let $X$ and $Y$ be independent exponential random variables, with

$$f(xmidlambda)=frac1lambdaexpleft(-fracxlambdaright),,x>0,, qquad f(ymidmu)=frac1muexpleft(-fracymuright),,y>0$$

We observe $Z$ and $W$ with $Z=min(X,Y)$, and $W=begincases 1 &,textif Z=X\ 0 &,textif Z=Y endcases$

I have obtained the joint distribution of $Z$ and $W$, i.e.,
$$P(Z leq z, W=0)=fraclambdamu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

$$P(Z leq z, W=1)=fracmumu+lambdaleft[1-expleft(-left(frac1mu+frac1lambdaright)zright)right]$$

Now assume that $(Z_i,W_i),i=1,cdots,n$, are $n$ i.i.d observations. Find the MLEs of $lambda$ and $mu$.

(This is the exercise 7.14 of the book Statistical Inference 2nd edition, but no solution given)

statistics maximum-likelihood exponential-distribution censoring

share|cite|improve this question

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

$endgroup$

share|cite|improve this question

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

share|cite|improve this question

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

edited Mar 21 at 10:03

StubbornAtom

6,29831440

edited Mar 21 at 10:03

StubbornAtom

6,29831440

asked Mar 21 at 7:16

Tim XuTim Xu

63

asked Mar 21 at 7:16

Tim XuTim Xu

63

2 Answers
2

active

oldest

votes

1

$begingroup$

Note that $Z$ and $W$ are in fact independent.

For $z>0,,,win0,1$, you can write the joint distribution of $(Z,W)$ as

$$P(z,w)=frac1lambda^w mu^1-wexpleft[-left(frac1lambda+frac1muright)zright]$$

The above formulation is not completely rigorous as it is not a probability (I am searching for a better notation avoiding the dirac delta).

The likelihood function given the sample $(z_1,w_1),ldots,(z_n,w_n)$ is then

$$L(lambda,mu)=frac1lambda^sum_i=1^n w_imu^n-sum_i=1^n w_iexpleft[-left(frac1lambda+frac1muright)sum_i=1^n z_iright]$$

Log-likelihood is

$$ell(lambda,mu)=-sum_i=1^n w_ilnlambda-left(n-sum_i=1^n w_iright)lnmu--left(frac1lambda+frac1muright)sum_i=1^n z_i$$

For $0<bar w<1$, solving for the stationary points of $ell(lambda,mu)$ yields $$hatlambda=fracsum_i=1^n z_isum_i=1^n w_i=fracbar zbar wqquad,qquad hatmu=fracsum_i=1^n z_in-sum_i=1^n w_i=fracbar z1-bar w$$

So assuming $0<bar w<1$, the unique MLE of $(lambda,mu)$ is $(hatlambda,hatmu)$.

But when $bar win0,1$, the MLE does not exist.

share|cite|improve this answer

edited Mar 21 at 12:29

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still have two questions. (Q1) How to solve for the stationary points of $ell(lambda,mu)$? (Q2) Why do we need to discuss the value range of $barw$? (In addition) Shouldn't the $mu$ in your Log-likelihood is $lnmu$?
  $endgroup$
  – Tim Xu
  Mar 21 at 12:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the log-likelihood. I will edit.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TimXu I just differentiated $ell$ once wrt $lambda$ and $mu$ and set them equal to zero (for multiple parameters, one should ideally check whether the Hessian matrix of the log-likelihood is negative definite at $(hatlambda,hatmu)$, provided the derivatives exist). And you can see from the expression for $hatlambda,hatmu$ that they are not defined at $bar w=0,1$. Hence the range distinction.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:35
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The $W=0$ and the $W=1$ case can be combined by writing $lambda^1-Wmu^W$.

First differentiating w.r.t. $z$ and forming the product the likelihood can be written as $prod_i^nfrac1lambda^w_imu^1-w_ie^-(frac1lambda+frac1mu)z_i$.
Taking logs gives the log likelihood $=-sum_i^n(w_ilnlambda+(1-w_i)lnmu+(frac1lambda+frac1mu)z_i)$.

Maximising wrt $lambda$ and $mu$ gives $lambda=fracbarzbarw$ and $mu=fracbarz(1-barw)$

Note if $lambda=mu$ then $barwapproxfrac12$ so the estimates for $lambda$ and $mu$ become equal.

share|cite|improve this answer

answered Mar 21 at 8:40

user121049user121049

1,360174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Denote the log likelihood by $log L$. First differentiating w.r.t. $lambda$ and $mu$, gives $hatlambda=fracbarzbarw$, $hatmu=fracbarz1-barw$. To make sure they are the maximizers of the log likelihood. I calculate their second-order differentials w.r.t. $lambda$ and $mu$, and get that $fracpartial^2 log Lpartial lambda^2=-sum_i=1^n(-fracw_ilambda^2+2fracz_ilambda^3)$ and $fracpartial^2 log Lpartial mu^2=-sum_i=1^n(-frac1-w_imu^2+2fracz_imu^3)$. How do you know these two second-order differentials are negative?
  $endgroup$
  – Tim Xu
  Mar 21 at 11:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Direct substitution. for instance $fracpartial^2 logLpartial lambda^2=-fracbarw^3barz^2$
  $endgroup$
  – user121049
  Mar 21 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You missed an $n$. It is $fracpartial ^2 logLpartial lambda^2 |_lambda=fracbar zbar w= -sum_i=1^n (-fracw_ilambda^2+2fracz_ilambda^3) |_lambda=fracbar zbar w= (fracsum_i=1^n w_ilambda^2-2fracsum_i=1^n z_ilambda^3) |_lambda=fracbar zbar w= (fracn bar wlambda^2-2fracn bar zlambda^3) |_lambda=fracbar zbar w= fracn bar w^3bar z^2-2fracn bar z bar w^3bar z^3= fracn bar w^3bar z^2-2fracn bar w^3bar z^2=-fracn bar w^3bar z^2$
  $endgroup$
  – Tim Xu
  Mar 22 at 6:41
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Note that $Z$ and $W$ are in fact independent.

For $z>0,,,win0,1$, you can write the joint distribution of $(Z,W)$ as

$$P(z,w)=frac1lambda^w mu^1-wexpleft[-left(frac1lambda+frac1muright)zright]$$

The above formulation is not completely rigorous as it is not a probability (I am searching for a better notation avoiding the dirac delta).

The likelihood function given the sample $(z_1,w_1),ldots,(z_n,w_n)$ is then

$$L(lambda,mu)=frac1lambda^sum_i=1^n w_imu^n-sum_i=1^n w_iexpleft[-left(frac1lambda+frac1muright)sum_i=1^n z_iright]$$

Log-likelihood is

$$ell(lambda,mu)=-sum_i=1^n w_ilnlambda-left(n-sum_i=1^n w_iright)lnmu--left(frac1lambda+frac1muright)sum_i=1^n z_i$$

For $0<bar w<1$, solving for the stationary points of $ell(lambda,mu)$ yields $$hatlambda=fracsum_i=1^n z_isum_i=1^n w_i=fracbar zbar wqquad,qquad hatmu=fracsum_i=1^n z_in-sum_i=1^n w_i=fracbar z1-bar w$$

So assuming $0<bar w<1$, the unique MLE of $(lambda,mu)$ is $(hatlambda,hatmu)$.

But when $bar win0,1$, the MLE does not exist.

share|cite|improve this answer

edited Mar 21 at 12:29

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still have two questions. (Q1) How to solve for the stationary points of $ell(lambda,mu)$? (Q2) Why do we need to discuss the value range of $barw$? (In addition) Shouldn't the $mu$ in your Log-likelihood is $lnmu$?
  $endgroup$
  – Tim Xu
  Mar 21 at 12:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the log-likelihood. I will edit.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TimXu I just differentiated $ell$ once wrt $lambda$ and $mu$ and set them equal to zero (for multiple parameters, one should ideally check whether the Hessian matrix of the log-likelihood is negative definite at $(hatlambda,hatmu)$, provided the derivatives exist). And you can see from the expression for $hatlambda,hatmu$ that they are not defined at $bar w=0,1$. Hence the range distinction.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:35
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Note that $Z$ and $W$ are in fact independent.

For $z>0,,,win0,1$, you can write the joint distribution of $(Z,W)$ as

$$P(z,w)=frac1lambda^w mu^1-wexpleft[-left(frac1lambda+frac1muright)zright]$$

The above formulation is not completely rigorous as it is not a probability (I am searching for a better notation avoiding the dirac delta).

The likelihood function given the sample $(z_1,w_1),ldots,(z_n,w_n)$ is then

$$L(lambda,mu)=frac1lambda^sum_i=1^n w_imu^n-sum_i=1^n w_iexpleft[-left(frac1lambda+frac1muright)sum_i=1^n z_iright]$$

Log-likelihood is

$$ell(lambda,mu)=-sum_i=1^n w_ilnlambda-left(n-sum_i=1^n w_iright)lnmu--left(frac1lambda+frac1muright)sum_i=1^n z_i$$

For $0<bar w<1$, solving for the stationary points of $ell(lambda,mu)$ yields $$hatlambda=fracsum_i=1^n z_isum_i=1^n w_i=fracbar zbar wqquad,qquad hatmu=fracsum_i=1^n z_in-sum_i=1^n w_i=fracbar z1-bar w$$

So assuming $0<bar w<1$, the unique MLE of $(lambda,mu)$ is $(hatlambda,hatmu)$.

But when $bar win0,1$, the MLE does not exist.

share|cite|improve this answer

edited Mar 21 at 12:29

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I still have two questions. (Q1) How to solve for the stationary points of $ell(lambda,mu)$? (Q2) Why do we need to discuss the value range of $barw$? (In addition) Shouldn't the $mu$ in your Log-likelihood is $lnmu$?
  $endgroup$
  – Tim Xu
  Mar 21 at 12:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about the log-likelihood. I will edit.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TimXu I just differentiated $ell$ once wrt $lambda$ and $mu$ and set them equal to zero (for multiple parameters, one should ideally check whether the Hessian matrix of the log-likelihood is negative definite at $(hatlambda,hatmu)$, provided the derivatives exist). And you can see from the expression for $hatlambda,hatmu$ that they are not defined at $bar w=0,1$. Hence the range distinction.
  $endgroup$
  – StubbornAtom
  Mar 21 at 12:35
  
  

  
  
  
  

add a comment | 

$begingroup$

Note that $Z$ and $W$ are in fact independent.

For $z>0,,,win0,1$, you can write the joint distribution of $(Z,W)$ as

$$P(z,w)=frac1lambda^w mu^1-wexpleft[-left(frac1lambda+frac1muright)zright]$$

The above formulation is not completely rigorous as it is not a probability (I am searching for a better notation avoiding the dirac delta).

The likelihood function given the sample $(z_1,w_1),ldots,(z_n,w_n)$ is then

$$L(lambda,mu)=frac1lambda^sum_i=1^n w_imu^n-sum_i=1^n w_iexpleft[-left(frac1lambda+frac1muright)sum_i=1^n z_iright]$$

Log-likelihood is

$$ell(lambda,mu)=-sum_i=1^n w_ilnlambda-left(n-sum_i=1^n w_iright)lnmu--left(frac1lambda+frac1muright)sum_i=1^n z_i$$

For $0<bar w<1$, solving for the stationary points of $ell(lambda,mu)$ yields $$hatlambda=fracsum_i=1^n z_isum_i=1^n w_i=fracbar zbar wqquad,qquad hatmu=fracsum_i=1^n z_in-sum_i=1^n w_i=fracbar z1-bar w$$

So assuming $0<bar w<1$, the unique MLE of $(lambda,mu)$ is $(hatlambda,hatmu)$.

But when $bar win0,1$, the MLE does not exist.

share|cite|improve this answer

edited Mar 21 at 12:29

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

$endgroup$

share|cite|improve this answer

edited Mar 21 at 12:29

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

answered Mar 21 at 11:41

StubbornAtomStubbornAtom

6,29831440

0

$begingroup$

The $W=0$ and the $W=1$ case can be combined by writing $lambda^1-Wmu^W$.

First differentiating w.r.t. $z$ and forming the product the likelihood can be written as $prod_i^nfrac1lambda^w_imu^1-w_ie^-(frac1lambda+frac1mu)z_i$.
Taking logs gives the log likelihood $=-sum_i^n(w_ilnlambda+(1-w_i)lnmu+(frac1lambda+frac1mu)z_i)$.

Maximising wrt $lambda$ and $mu$ gives $lambda=fracbarzbarw$ and $mu=fracbarz(1-barw)$

Note if $lambda=mu$ then $barwapproxfrac12$ so the estimates for $lambda$ and $mu$ become equal.

share|cite|improve this answer

answered Mar 21 at 8:40

user121049user121049

1,360174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Denote the log likelihood by $log L$. First differentiating w.r.t. $lambda$ and $mu$, gives $hatlambda=fracbarzbarw$, $hatmu=fracbarz1-barw$. To make sure they are the maximizers of the log likelihood. I calculate their second-order differentials w.r.t. $lambda$ and $mu$, and get that $fracpartial^2 log Lpartial lambda^2=-sum_i=1^n(-fracw_ilambda^2+2fracz_ilambda^3)$ and $fracpartial^2 log Lpartial mu^2=-sum_i=1^n(-frac1-w_imu^2+2fracz_imu^3)$. How do you know these two second-order differentials are negative?
  $endgroup$
  – Tim Xu
  Mar 21 at 11:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Direct substitution. for instance $fracpartial^2 logLpartial lambda^2=-fracbarw^3barz^2$
  $endgroup$
  – user121049
  Mar 21 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You missed an $n$. It is $fracpartial ^2 logLpartial lambda^2 |_lambda=fracbar zbar w= -sum_i=1^n (-fracw_ilambda^2+2fracz_ilambda^3) |_lambda=fracbar zbar w= (fracsum_i=1^n w_ilambda^2-2fracsum_i=1^n z_ilambda^3) |_lambda=fracbar zbar w= (fracn bar wlambda^2-2fracn bar zlambda^3) |_lambda=fracbar zbar w= fracn bar w^3bar z^2-2fracn bar z bar w^3bar z^3= fracn bar w^3bar z^2-2fracn bar w^3bar z^2=-fracn bar w^3bar z^2$
  $endgroup$
  – Tim Xu
  Mar 22 at 6:41
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The $W=0$ and the $W=1$ case can be combined by writing $lambda^1-Wmu^W$.

First differentiating w.r.t. $z$ and forming the product the likelihood can be written as $prod_i^nfrac1lambda^w_imu^1-w_ie^-(frac1lambda+frac1mu)z_i$.
Taking logs gives the log likelihood $=-sum_i^n(w_ilnlambda+(1-w_i)lnmu+(frac1lambda+frac1mu)z_i)$.

Maximising wrt $lambda$ and $mu$ gives $lambda=fracbarzbarw$ and $mu=fracbarz(1-barw)$

Note if $lambda=mu$ then $barwapproxfrac12$ so the estimates for $lambda$ and $mu$ become equal.

share|cite|improve this answer

answered Mar 21 at 8:40

user121049user121049

1,360174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Denote the log likelihood by $log L$. First differentiating w.r.t. $lambda$ and $mu$, gives $hatlambda=fracbarzbarw$, $hatmu=fracbarz1-barw$. To make sure they are the maximizers of the log likelihood. I calculate their second-order differentials w.r.t. $lambda$ and $mu$, and get that $fracpartial^2 log Lpartial lambda^2=-sum_i=1^n(-fracw_ilambda^2+2fracz_ilambda^3)$ and $fracpartial^2 log Lpartial mu^2=-sum_i=1^n(-frac1-w_imu^2+2fracz_imu^3)$. How do you know these two second-order differentials are negative?
  $endgroup$
  – Tim Xu
  Mar 21 at 11:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Direct substitution. for instance $fracpartial^2 logLpartial lambda^2=-fracbarw^3barz^2$
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  – user121049
  Mar 21 at 14:09
  
  

  
  
  
  


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  You missed an $n$. It is $fracpartial ^2 logLpartial lambda^2 |_lambda=fracbar zbar w= -sum_i=1^n (-fracw_ilambda^2+2fracz_ilambda^3) |_lambda=fracbar zbar w= (fracsum_i=1^n w_ilambda^2-2fracsum_i=1^n z_ilambda^3) |_lambda=fracbar zbar w= (fracn bar wlambda^2-2fracn bar zlambda^3) |_lambda=fracbar zbar w= fracn bar w^3bar z^2-2fracn bar z bar w^3bar z^3= fracn bar w^3bar z^2-2fracn bar w^3bar z^2=-fracn bar w^3bar z^2$
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  – Tim Xu
  Mar 22 at 6:41
  
  

  
  
  
  

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The $W=0$ and the $W=1$ case can be combined by writing $lambda^1-Wmu^W$.

First differentiating w.r.t. $z$ and forming the product the likelihood can be written as $prod_i^nfrac1lambda^w_imu^1-w_ie^-(frac1lambda+frac1mu)z_i$.
Taking logs gives the log likelihood $=-sum_i^n(w_ilnlambda+(1-w_i)lnmu+(frac1lambda+frac1mu)z_i)$.

Maximising wrt $lambda$ and $mu$ gives $lambda=fracbarzbarw$ and $mu=fracbarz(1-barw)$

Note if $lambda=mu$ then $barwapproxfrac12$ so the estimates for $lambda$ and $mu$ become equal.

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answered Mar 21 at 8:40

user121049user121049

1,360174

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share|cite|improve this answer

answered Mar 21 at 8:40

user121049user121049

1,360174

answered Mar 21 at 8:40

user121049user121049

1,360174

answered Mar 21 at 8:40

user121049user121049

1,360174