Mixing
Let $X$ be Poisson r.v. with $lambda$ find $f(x)$ such that $E[f(X)]=lambda log (lambda)$
$begingroup$
I am looking for a function $f(x)$ such that
begin{align}
E[f(X)]=lambda log (lambda) quad text{ for all } lambda ge 0 quad (*)
end{align}
where $X$ is a Poisson random varaible with parameter $lambda$. Note, we are looking for a function idenpendent of $lambda$.
Here are some thoughts:
begin{align}
lambda log (lambda) = E[f(X)]= sum_{k=0}^infty f(k) frac{lambda^k e^{-lambda}}{k!}
end{align}
Therefore, we have that
begin{align}
e^{lambda} lambda log (lambda) = sum_{k=0}^infty f(k) frac{lambda^k }{k!}= sum_{k=0}^infty a_k frac{lambda^k }{k!} quad (**)
end{align}
where in the last step I defined $f(k)=a_k$.
Now, this is the point where I am a bit stuck.
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x log(x)$ can be written as a Maclaurin series.
These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.
real-analysis probability probability-theory poisson-distribution
asked Jan 13 at 21:25
BobyBoby
1,0511929
$endgroup$
add a comment |
$begingroup$
I am looking for a function $f(x)$ such that
begin{align}
E[f(X)]=lambda log (lambda) quad text{ for all } lambda ge 0 quad (*)
end{align}
where $X$ is a Poisson random varaible with parameter $lambda$. Note, we are looking for a function idenpendent of $lambda$.
Here are some thoughts:
begin{align}
lambda log (lambda) = E[f(X)]= sum_{k=0}^infty f(k) frac{lambda^k e^{-lambda}}{k!}
end{align}
Therefore, we have that
begin{align}
e^{lambda} lambda log (lambda) = sum_{k=0}^infty f(k) frac{lambda^k }{k!}= sum_{k=0}^infty a_k frac{lambda^k }{k!} quad (**)
end{align}
where in the last step I defined $f(k)=a_k$.
Now, this is the point where I am a bit stuck.
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x log(x)$ can be written as a Maclaurin series.
These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.
real-analysis probability probability-theory poisson-distribution
asked Jan 13 at 21:25
BobyBoby
1,0511929
$endgroup$
2
$begingroup$
Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
$endgroup$
– LoveTooNap29
Jan 13 at 21:30
$begingroup$
@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
$endgroup$
– Boby
Jan 13 at 21:32
add a comment |
$begingroup$
I am looking for a function $f(x)$ such that
begin{align}
E[f(X)]=lambda log (lambda) quad text{ for all } lambda ge 0 quad (*)
end{align}
where $X$ is a Poisson random varaible with parameter $lambda$. Note, we are looking for a function idenpendent of $lambda$.
Here are some thoughts:
begin{align}
lambda log (lambda) = E[f(X)]= sum_{k=0}^infty f(k) frac{lambda^k e^{-lambda}}{k!}
end{align}
Therefore, we have that
begin{align}
e^{lambda} lambda log (lambda) = sum_{k=0}^infty f(k) frac{lambda^k }{k!}= sum_{k=0}^infty a_k frac{lambda^k }{k!} quad (**)
end{align}
where in the last step I defined $f(k)=a_k$.
Now, this is the point where I am a bit stuck.
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x log(x)$ can be written as a Maclaurin series.
These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.
real-analysis probability probability-theory poisson-distribution
asked Jan 13 at 21:25
BobyBoby
1,0511929
$endgroup$
I am looking for a function $f(x)$ such that
begin{align}
E[f(X)]=lambda log (lambda) quad text{ for all } lambda ge 0 quad (*)
end{align}
where $X$ is a Poisson random varaible with parameter $lambda$. Note, we are looking for a function idenpendent of $lambda$.
Here are some thoughts:
begin{align}
lambda log (lambda) = E[f(X)]= sum_{k=0}^infty f(k) frac{lambda^k e^{-lambda}}{k!}
end{align}
Therefore, we have that
begin{align}
e^{lambda} lambda log (lambda) = sum_{k=0}^infty f(k) frac{lambda^k }{k!}= sum_{k=0}^infty a_k frac{lambda^k }{k!} quad (**)
end{align}
where in the last step I defined $f(k)=a_k$.
Now, this is the point where I am a bit stuck.
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x log(x)$ can be written as a Maclaurin series.
These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.
real-analysis probability probability-theory poisson-distribution
real-analysis probability probability-theory poisson-distribution
asked Jan 13 at 21:25
BobyBoby
1,0511929
asked Jan 13 at 21:25
BobyBoby
1,0511929
edited Jan 13 at 22:28
Boby
asked Jan 13 at 21:25
BobyBoby
1,0511929
asked Jan 13 at 21:25
BobyBoby
1,0511929
asked Jan 13 at 21:25
BobyBoby
1,0511929
1,0511929
2
$begingroup$
Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
$endgroup$
– LoveTooNap29
Jan 13 at 21:30
$begingroup$
@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
$endgroup$
– Boby
Jan 13 at 21:32
add a comment |
2
$begingroup$
Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
$endgroup$
– LoveTooNap29
Jan 13 at 21:30
$begingroup$
@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
$endgroup$
– Boby
Jan 13 at 21:32
2
2
$begingroup$
Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
$endgroup$
– LoveTooNap29
Jan 13 at 21:30
$begingroup$
Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
$endgroup$
– LoveTooNap29
Jan 13 at 21:30
$begingroup$
@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
$endgroup$
– Boby
Jan 13 at 21:32
$begingroup$
@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
$endgroup$
– Boby
Jan 13 at 21:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To be precise, we want to find $f:mathbb{N}_0tomathbb{C}$ such that
$
Eleft[|f(X)|right]<infty,
$ and $$
E[f(X)]=lambdaloglambda,quadforall lambdage 0.
$$
Now, we can write
$$
sum_{k=0}^infty f(k)frac{lambda^k e^{-lambda}}{k!}=lambdaloglambda,
$$or equivalently
$$
sum_{k=0}^infty f(k)frac{lambda^k}{k!}=lambda e^lambdaloglambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $lambdage 0$. Then this implies that it can be extended uniquely to some entire function $F(lambda)$. Since $F(0) = limlimits_{lambdato 0^+}lambda e^lambda log lambda =0$, $G(lambda)=frac{F(lambda)} {lambda}$ also defines an entire function. Now, we have
$$
e^{-lambda}G(lambda) =loglambda
$$ for all $lambda>0$, but $limlimits_{lambdato 0^+}e^{-lambda}G(lambda)=G(0)neq limlimits_{lambdato 0^+}loglambda=-infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<infty$ and $E[f(X)]=lambda log lambda$ for all $lambdage 0$.
answered Jan 13 at 21:54
SongSong
13.1k632
$endgroup$
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
add a comment |
Your Answer
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1 Answer
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$begingroup$
To be precise, we want to find $f:mathbb{N}_0tomathbb{C}$ such that
$
Eleft[|f(X)|right]<infty,
$ and $$
E[f(X)]=lambdaloglambda,quadforall lambdage 0.
$$
Now, we can write
$$
sum_{k=0}^infty f(k)frac{lambda^k e^{-lambda}}{k!}=lambdaloglambda,
$$or equivalently
$$
sum_{k=0}^infty f(k)frac{lambda^k}{k!}=lambda e^lambdaloglambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $lambdage 0$. Then this implies that it can be extended uniquely to some entire function $F(lambda)$. Since $F(0) = limlimits_{lambdato 0^+}lambda e^lambda log lambda =0$, $G(lambda)=frac{F(lambda)} {lambda}$ also defines an entire function. Now, we have
$$
e^{-lambda}G(lambda) =loglambda
$$ for all $lambda>0$, but $limlimits_{lambdato 0^+}e^{-lambda}G(lambda)=G(0)neq limlimits_{lambdato 0^+}loglambda=-infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<infty$ and $E[f(X)]=lambda log lambda$ for all $lambdage 0$.
answered Jan 13 at 21:54
SongSong
13.1k632
$endgroup$
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
add a comment |
$begingroup$
To be precise, we want to find $f:mathbb{N}_0tomathbb{C}$ such that
$
Eleft[|f(X)|right]<infty,
$ and $$
E[f(X)]=lambdaloglambda,quadforall lambdage 0.
$$
Now, we can write
$$
sum_{k=0}^infty f(k)frac{lambda^k e^{-lambda}}{k!}=lambdaloglambda,
$$or equivalently
$$
sum_{k=0}^infty f(k)frac{lambda^k}{k!}=lambda e^lambdaloglambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $lambdage 0$. Then this implies that it can be extended uniquely to some entire function $F(lambda)$. Since $F(0) = limlimits_{lambdato 0^+}lambda e^lambda log lambda =0$, $G(lambda)=frac{F(lambda)} {lambda}$ also defines an entire function. Now, we have
$$
e^{-lambda}G(lambda) =loglambda
$$ for all $lambda>0$, but $limlimits_{lambdato 0^+}e^{-lambda}G(lambda)=G(0)neq limlimits_{lambdato 0^+}loglambda=-infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<infty$ and $E[f(X)]=lambda log lambda$ for all $lambdage 0$.
answered Jan 13 at 21:54
SongSong
13.1k632
$endgroup$
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
add a comment |
$begingroup$
To be precise, we want to find $f:mathbb{N}_0tomathbb{C}$ such that
$
Eleft[|f(X)|right]<infty,
$ and $$
E[f(X)]=lambdaloglambda,quadforall lambdage 0.
$$
Now, we can write
$$
sum_{k=0}^infty f(k)frac{lambda^k e^{-lambda}}{k!}=lambdaloglambda,
$$or equivalently
$$
sum_{k=0}^infty f(k)frac{lambda^k}{k!}=lambda e^lambdaloglambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $lambdage 0$. Then this implies that it can be extended uniquely to some entire function $F(lambda)$. Since $F(0) = limlimits_{lambdato 0^+}lambda e^lambda log lambda =0$, $G(lambda)=frac{F(lambda)} {lambda}$ also defines an entire function. Now, we have
$$
e^{-lambda}G(lambda) =loglambda
$$ for all $lambda>0$, but $limlimits_{lambdato 0^+}e^{-lambda}G(lambda)=G(0)neq limlimits_{lambdato 0^+}loglambda=-infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<infty$ and $E[f(X)]=lambda log lambda$ for all $lambdage 0$.
answered Jan 13 at 21:54
SongSong
13.1k632
$endgroup$
To be precise, we want to find $f:mathbb{N}_0tomathbb{C}$ such that
$
Eleft[|f(X)|right]<infty,
$ and $$
E[f(X)]=lambdaloglambda,quadforall lambdage 0.
$$
Now, we can write
$$
sum_{k=0}^infty f(k)frac{lambda^k e^{-lambda}}{k!}=lambdaloglambda,
$$or equivalently
$$
sum_{k=0}^infty f(k)frac{lambda^k}{k!}=lambda e^lambdaloglambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $lambdage 0$. Then this implies that it can be extended uniquely to some entire function $F(lambda)$. Since $F(0) = limlimits_{lambdato 0^+}lambda e^lambda log lambda =0$, $G(lambda)=frac{F(lambda)} {lambda}$ also defines an entire function. Now, we have
$$
e^{-lambda}G(lambda) =loglambda
$$ for all $lambda>0$, but $limlimits_{lambdato 0^+}e^{-lambda}G(lambda)=G(0)neq limlimits_{lambdato 0^+}loglambda=-infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<infty$ and $E[f(X)]=lambda log lambda$ for all $lambdage 0$.
answered Jan 13 at 21:54
SongSong
13.1k632
answered Jan 13 at 21:54
SongSong
13.1k632
answered Jan 13 at 21:54
SongSong
13.1k632
answered Jan 13 at 21:54
SongSong
13.1k632
13.1k632
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
add a comment |
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
Thanks. Why does $G(0)neq -infty$? Is it because otherwise, it is equal to zero at that point?
$endgroup$
– Boby
Jan 13 at 22:01
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
@Boby That is because $G(z)=frac{F(z)}{z}$ must have a removable singularity at $z=0$ (note that $F$ is already an entire function vanishing at $0$) and hence $G$ is also entire.
$endgroup$
– Song
Jan 13 at 22:06
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
$begingroup$
Also, how important is the assumption $E[|f(X)|]<infty$ in this proof? Can there exists a function such that that $E[f(X)]=lambda log(lambda)$ while $E[|f(X)|]=infty$ ?
$endgroup$
– Boby
Jan 13 at 22:16
1
1
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
$begingroup$
Expectation is 'defined' only for those $f$ such that $Elvert f(X)rvert <infty.$ Otherwise, expected value is not even defined and there is nothing we can say about it. So it is a matter of the definition of an expecation (or integral with respect to a measure).
$endgroup$
– Song
Jan 13 at 22:23
add a comment |
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Well, by linearity, $mathbb{E}(log(lambda ) X)=log lambda mathbb{E}(X)=lambda log lambda$, since $log lambda$ is a constant, yes? So, you might need an extra condition to avoid such trivial answers of $f(x)=log(lambda) cdot x$
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– LoveTooNap29
Jan 13 at 21:30
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@LoveTooNap29 Yeah, I want a function independent of $lambda$. Let me add this. Thank you, this example did not occur to me.
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– Boby
Jan 13 at 21:32