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According to wikipedia location parameter is such a number $x_0$ that satisfies:

$$f_{x_0, \theta} (x)= f_\theta(x - x_0)$$

My main goal is to calculate second order Taylor expansion of probability mass function of poisson distribution around location parameter. To do this my first idea was just to derive form of this location parameter:

In our example location parameter $k_0$ will be a number that satisfies:

$$\frac{\lambda^ke^{-\lambda}}{k!} = \frac{\lambda^{k - k_0}e^{-\lambda}}{(k-k_0)!}$$

$$\frac{1}{k!} = \frac{\lambda^{-k_0}}{(k - k_0)!}$$

$$-\ln(k!)=-k_0 \ln(\lambda) - \ln[(k -k_0)!]$$

Let's rewrite $\ln(k!)$: $$\ln(k!) = \ln((k - k_0)! \cdot (k - k_0 + 1) \cdot k-k_0+2 \cdot ... \cdot k) = \ln(k - k_0) + \ln[\prod_{i = 1}^{k_0} k - k_0 + i]$$

Putting this together:

$$-\ln(k-k_0) - \ln[\prod_{i = 1}^{k_0} k - k_0 + i] = -k_0 \ln(\lambda) - ln[(k - k_0)!]$$

$$k_0 = \frac{\ln[\prod_{i = 1}^{k_0} k - k_0 + i]}{\ln\lambda}$$

My questions are:

  1. Are my calculations correct? Is my calculated parameter $k_0$ really a location parameter?

  2. Is it necessary to first derive location parameter for calculating Taylor expansion afterwards? Is there any way around?

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  • $\begingroup$ The Poisson distribution is discrete. In Taylor expanding you are probably assuming it is continuous. $\endgroup$ Commented Dec 27, 2021 at 11:37
  • $\begingroup$ There is no reason to expect that equation to have an integer solution which makes the concept of a location parameter not really relevant. More you seem to be actually trying to do is describe the fluctuations around the mean as being Gaussian-like, which can be done, but you first need to center and rescale so as to produce a small parameter ($\lambda^{-1/2}$ in this context) with which to expand. $\endgroup$
    – Ian
    Commented Dec 27, 2021 at 11:43
  • $\begingroup$ Nevertheless, is my derivation correct? $\endgroup$
    – John
    Commented Dec 27, 2021 at 14:39

1 Answer 1

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I think you misapply the characterization you cite on wiki, which I believe simply says that the location parameter is a horizontal shift--being thought of as a newly added parameter-- to a PMF or PDF. See the more general notion of group family. The location parameter doesn't need to "solve" anything.

Instead, what you are solving is when the Poisson PMF is equal at two different points in its support, namely $k$ and what you have labelled as $k-k_0$. Since the Poisson PMF is nonmonotonic (see pic below), there may be two points with equal mass, but since the distribution is discrete, this is not necessarily guaranteed for arbitrary $k$. Moreover, as $k$ gets large, the PMF monotonically tends toward zero, so sufficiently large $k$ will have no partner point with the same mass.

enter image description here

Not sure what you want to do with Taylor expansions...

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  • $\begingroup$ Hmmm Okay.. To summarize Am I correct saying that when I define my density as $f(k, \lambda) = \frac{\lambda^{k - k_0}e^{-\lambda}}{(k-k_0)!}$, in this case $k_0$ is a location parameter? $\endgroup$
    – John
    Commented Dec 27, 2021 at 15:28
  • $\begingroup$ I'm just struggling with applying definition of location parameter to Poisson distribution. How it should look like in this probability mass? $\endgroup$
    – John
    Commented Dec 27, 2021 at 15:33
  • $\begingroup$ Sure-as per the link I shared on group family, I think it makes sense in that case to refer to $k_0$ as a location parameter for a shifted Poisson having support $k\in\{k_0,k_0+1,...\}$ $\endgroup$ Commented Dec 27, 2021 at 17:41

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