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This follows on from my question here.

I have the following problem:

Let $Y_1, \dots, Y_n$ be a random sample from a Poisson distribution $\text{Pois}(\lambda)$. Recall, the $\text{Pois}(\lambda)$ distribution has the probability function $f_{\lambda}(y) = e^{-\lambda} \dfrac{\lambda^y}{y!}$, if $y = 0, 1, 2, 3, \dots$, and $\lambda > 0$.

(a) Show that $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i$ is a sufficient statistic for $\lambda$ using the Fisher-Neyman factorisation theorem.

(b) What is the distribution of $T(\mathbf{Y})$? Obtain this result directly using the definition of a sufficient statistic.

For (a), we have that $L(\lambda, \mathbf{y}) = \prod_{i = 1}^n e^{-\lambda}\dfrac{\lambda^{y_i}}{y_i!} = e^{-n \lambda} \dfrac{\lambda^{\sum_{i = 1}^n y_i}}{\prod_{i = 1}^n y_i!}$. So $T(\mathbf{y}) = \sum_{i = 1}^n y_i$, $g(t, \lambda) = e^{-n \lambda} \lambda^t$ and $h(\mathbf{y}) = \dfrac{1}{\prod_{i = 1}^n y_i!}$.

For (b), the solution is given as follows:

$$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$

$$P(\mathbf{Y} \mid T(\mathbf{Y})) = \dfrac{P(\mathbf{Y}, T(\mathbf{Y}))}{P(T(\mathbf{Y}))} = \dfrac{\prod_{i = 1}^n e^{-\lambda} \dfrac{\lambda^{Y_i}}{Y_i!}}{e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}} = \dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$$

How does the result $\dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$ tell us what distribution $T(\mathbf{Y})$ is? Looking at it, I don't understand what point the problem was trying to make. Yes, we derived this result using the definition of sufficient statistic, but $\dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$ seems like quite an ugly result, in that it does not seem very/immediately insightful as to what the distribution of $T(\mathbf{Y})$ is.

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    $\begingroup$ First, whatever you might feel about this result, the formula completely defines the distribution because it specifies a set of positive probabilities that sum to unity. To most people this is a beautiful result: see the definition of the multinomial distribution and bear in mind that $T = Y_1+Y_2+\cdots+Y_n$ in the formula. $\endgroup$
    – whuber
    Apr 7, 2022 at 12:58
  • $\begingroup$ @whuber "First, whatever you might feel about this result, it completely defines the distribution because it specifies a set of positive probabilities that sum to unity." How so? I don't see this. $\endgroup$ Apr 7, 2022 at 13:01
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    $\begingroup$ It explicitly gives a formula for the probability of any outcome $\mathbf Y$ conditional on the sum $T(\mathbf Y).$ (Notice it was earlier specified that the components of $\mathbf Y$ can only be natural numbers $0, 1, 2, \ldots.$) $\endgroup$
    – whuber
    Apr 7, 2022 at 13:48
  • $\begingroup$ The question "What is the distribution of T(Y)?" is not difficult. But what is meant by "Obtain this result directly using the definition of a sufficient statistic" is not very clear. (I always hated these types of exam questions where the main problem is the understanding what is meant by the question). The answer that is given plainly starts with $$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$ so this is sort of begging the question. $\endgroup$ Apr 9, 2022 at 16:59

2 Answers 2

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Here is a slide from my mathematical statistics course that can help:

enter image description here

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https://en.wikipedia.org/wiki/Sufficient_statistic#Mathematical_definition

A statistic $t = T(Y)$ is sufficient for underlying parameter $\theta$ precisely if the conditional probability distribution of the data $Y$, given the statistic $t = T(Y)$, does not depend on the parameter $\theta$.

Thus we can describe

$$\overbrace{f(Y=y|T=t,\theta)}^{\text{distribution of the data $X$ given $T = t$, and $\theta$}} = \overbrace{g(Y=y|T=t)}^{\text{function without $\theta$}}$$

Then we can apply that the joint distribution is equal to the conditional distribution multiplied with the distribution of the condition

$$\begin{array}{rcccl}f(Y=y \text{ and } T=t | \theta) &=& f(Y=y|T=t,\theta) &\cdot& f(T=t|\theta) \\ &=& g(Y=y|T=t) &\cdot& f(T=t|\theta) \\ &=& \dfrac{1}{n^t} \dfrac{t!}{\prod_{i = 1}^n y_i!} &\cdot& f(T=t|\theta) \end{array}$$

then

$$f(T = t |\theta) = \frac{f(Y=y \text{ and } T=t | \theta) }{\dfrac{1}{n^t} \dfrac{t!}{\prod_{i = 1}^n y_i!}} $$

and we can use any arbitrary sample $Y$ to compute the above expression on the right side. So let's do it easy and choose $y_1 = t$ while $y_i = 0$ for $i>1$. Then

$$\begin{array}{rcl}f(T = t |\theta) &=& \frac{f(Y_1=t, Y_2 = Y_3 = \dots = Y_n = 0 \text{ and } T=t | \theta) }{\dfrac{1}{n^t} \dfrac{t!}{t!}} \\ &=& \frac{\lambda^t e^{-\lambda}/t! \cdot \left( \lambda^0 e^{-\lambda}/0! \right)^{n-1}}{\dfrac{1}{n^t}}\\ & =&\frac{\lambda^t e^{-n\lambda}}{t!\dfrac{1}{n^t}} \\ & =&\frac{(n\lambda)^t e^{-n\lambda}}{t!} \end{array}$$

which is a Poisson distribution with rate parameter $n\lambda$.


Above, I derived the distribution of $T(Y)$ given the conditional distribution of $Y$ given $T(Y)$. I do not get in your question how the quote below is an answer to (b)

$$T(\mathbf{Y}) \sim \text{Pois}(n \lambda)$$

$$P(\mathbf{Y} \mid T(\mathbf{Y})) = \dfrac{P(\mathbf{Y}, T(\mathbf{Y}))}{P(T(\mathbf{Y}))} = \dfrac{\prod_{i = 1}^n e^{-\lambda} \dfrac{\lambda^{Y_i}}{Y_i!}}{e^{-n \lambda}\dfrac{n^T \lambda^T}{T!}} = \dfrac{1}{n^T} \dfrac{T!}{\prod_{i = 1}^n Y_i!}$$

That answer derives $P(\mathbf{Y} \mid T(\mathbf{Y}))$ instead of $P(T(\mathbf{Y}))$

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