Mixing
MLE of $lambda$ Given $f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x}$
$begingroup$
$f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $
$0$ otherwise
What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$
My input
Liklihood function of the sample is given by
$L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$
=$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
Am I following correct path ? Can someone tell me ?
statistical-inference maximum-likelihood
asked Jan 10 at 12:04
Daman deepDaman deep
751318
$endgroup$
add a comment |
$begingroup$
$f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $
$0$ otherwise
What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$
My input
Liklihood function of the sample is given by
$L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$
=$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
Am I following correct path ? Can someone tell me ?
statistical-inference maximum-likelihood
asked Jan 10 at 12:04
Daman deepDaman deep
751318
$endgroup$
add a comment |
$begingroup$
$f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $
$0$ otherwise
What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$
My input
Liklihood function of the sample is given by
$L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$
=$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
Am I following correct path ? Can someone tell me ?
statistical-inference maximum-likelihood
asked Jan 10 at 12:04
Daman deepDaman deep
751318
$endgroup$
$f(x;lambda)=1-dfrac{2}{3}lambda+lambdasqrt{x} ; 0le xle1 $
$0$ otherwise
What is the maximum likelihood estimate of the parameter $lambda$ based on two independent observations $x_1=dfrac{1}{4}$ and $x_2=dfrac{9}{16}$
My input
Liklihood function of the sample is given by
$L(lambda)=prod_{i=1}^{2}f(x_i;lambda)=(1-dfrac{2}{3}lambda+lambdasqrt{x_i})$
=$(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$log(L(lambda)=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
$=log(1-dfrac{2}{3}lambda+lambdasqrt{x_1})+log(1-dfrac{2}{3}lambda+lambdasqrt{x_2})$
Am I following correct path ? Can someone tell me ?
statistical-inference maximum-likelihood
statistical-inference maximum-likelihood
asked Jan 10 at 12:04
Daman deepDaman deep
751318
asked Jan 10 at 12:04
Daman deepDaman deep
751318
asked Jan 10 at 12:04
Daman deepDaman deep
751318
asked Jan 10 at 12:04
Daman deepDaman deep
751318
asked Jan 10 at 12:04
Daman deepDaman deep
751318
751318
add a comment |
add a comment |
1 Answer
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You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
add a comment |
$begingroup$
You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
add a comment |
$begingroup$
You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
$endgroup$
You are correct, but in this case it is easier to optimize the likelihood instead of the log-likelihood as it is a polynomial. So begin{align}L(lambda)&=left(1+left(sqrt{x_1}-frac23right)lambdaright)left(1+left(sqrt{x_2}-frac23right)lambdaright)\&=left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)lambda^2+left(sqrt{x_1}+sqrt{x_2}-frac43right)lambda+1\ implies L'left(hatlambdaright)&=2left(sqrt{x_1}-frac23right)left(sqrt{x_2}-frac23right)hatlambda+sqrt{x_1}+sqrt{x_2}-frac43=0end{align} There you can simply solve for $hatlambda$ and check whether it is a maximum or minimum.
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
answered Jan 10 at 13:15
TheSimpliFireTheSimpliFire
12.4k62460
12.4k62460
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
add a comment |
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
$hat lambda =8 $ But my lambda coming out to be $3 $ can you check if its coming out $8$ or not ?
$endgroup$
– Daman deep
Jan 10 at 13:22
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
Even to optimize a polynomial, especially expressed as a product, using logarithms is often the way to go.
$endgroup$
– Did
Jan 10 at 13:32
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
$begingroup$
@Did I suppose for quadratics and cubics then; once you differentiate them they are easy to factor
$endgroup$
– TheSimpliFire
Jan 10 at 14:16
add a comment |
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