Maximum likelihood estimation in a Poisson convolution












0














Suppose that $X$ and $Y$ are i.i.d. Poisson random variables, with mean $nu$. The parameter $nu$ is unknown and we would like to estimate it. We only are given the single data point
$$
X-Y.
$$

What is the maximum likelihood estimator for $nu$?



I am unsure how to go about this, or even if it is analytically tractable, because I cannot calculate a simple expression for
$$
mathbb{P}[X-Y=k].
$$

Is it in fact possible to find a simple form for $mathbb{P}[X-Y=k]$ and then maximise this with respect to $nu$? Or a more general question: is it always necessary to explicitly compute a probability distribution before determining a maximum likelihood estimator?



Anyway, the most sensible estimator choice that I can naively envisage is
$$
widehat{nu}=(X-Y)^2,
$$

which is unbiased at least. Could this be the MLE?










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  • For the difference of independently distributed Poisson variables, see Skellam distribution.
    – StubbornAtom
    Nov 18 at 19:01
















0














Suppose that $X$ and $Y$ are i.i.d. Poisson random variables, with mean $nu$. The parameter $nu$ is unknown and we would like to estimate it. We only are given the single data point
$$
X-Y.
$$

What is the maximum likelihood estimator for $nu$?



I am unsure how to go about this, or even if it is analytically tractable, because I cannot calculate a simple expression for
$$
mathbb{P}[X-Y=k].
$$

Is it in fact possible to find a simple form for $mathbb{P}[X-Y=k]$ and then maximise this with respect to $nu$? Or a more general question: is it always necessary to explicitly compute a probability distribution before determining a maximum likelihood estimator?



Anyway, the most sensible estimator choice that I can naively envisage is
$$
widehat{nu}=(X-Y)^2,
$$

which is unbiased at least. Could this be the MLE?










share|cite|improve this question






















  • For the difference of independently distributed Poisson variables, see Skellam distribution.
    – StubbornAtom
    Nov 18 at 19:01














0












0








0


0





Suppose that $X$ and $Y$ are i.i.d. Poisson random variables, with mean $nu$. The parameter $nu$ is unknown and we would like to estimate it. We only are given the single data point
$$
X-Y.
$$

What is the maximum likelihood estimator for $nu$?



I am unsure how to go about this, or even if it is analytically tractable, because I cannot calculate a simple expression for
$$
mathbb{P}[X-Y=k].
$$

Is it in fact possible to find a simple form for $mathbb{P}[X-Y=k]$ and then maximise this with respect to $nu$? Or a more general question: is it always necessary to explicitly compute a probability distribution before determining a maximum likelihood estimator?



Anyway, the most sensible estimator choice that I can naively envisage is
$$
widehat{nu}=(X-Y)^2,
$$

which is unbiased at least. Could this be the MLE?










share|cite|improve this question













Suppose that $X$ and $Y$ are i.i.d. Poisson random variables, with mean $nu$. The parameter $nu$ is unknown and we would like to estimate it. We only are given the single data point
$$
X-Y.
$$

What is the maximum likelihood estimator for $nu$?



I am unsure how to go about this, or even if it is analytically tractable, because I cannot calculate a simple expression for
$$
mathbb{P}[X-Y=k].
$$

Is it in fact possible to find a simple form for $mathbb{P}[X-Y=k]$ and then maximise this with respect to $nu$? Or a more general question: is it always necessary to explicitly compute a probability distribution before determining a maximum likelihood estimator?



Anyway, the most sensible estimator choice that I can naively envisage is
$$
widehat{nu}=(X-Y)^2,
$$

which is unbiased at least. Could this be the MLE?







poisson-distribution maximum-likelihood parameter-estimation






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 18 at 15:58









user301395

1899




1899












  • For the difference of independently distributed Poisson variables, see Skellam distribution.
    – StubbornAtom
    Nov 18 at 19:01


















  • For the difference of independently distributed Poisson variables, see Skellam distribution.
    – StubbornAtom
    Nov 18 at 19:01
















For the difference of independently distributed Poisson variables, see Skellam distribution.
– StubbornAtom
Nov 18 at 19:01




For the difference of independently distributed Poisson variables, see Skellam distribution.
– StubbornAtom
Nov 18 at 19:01















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