# #StackBounty: #normal-distribution #poisson-distribution #likelihood #sufficient-statistics Finding the form \$g(T(mathbf{y}), lambda)…

### Bounty: 50

I’m studying some notes that present examples of sufficiency:

Let $$Y_1, dots, Y_n$$ be i.i.d. $$N(mu, sigma^2)$$. Note that $$sum_{i = 1}^n (y_i – mu)^2 = sum_{i = 1}^n (y_i – bar{y})^2 + n(bar{y} – mu)^2$$. Hence

begin{align} L(mu, sigma; mathbf{y}) &= prod_{i = 1}^n dfrac{1}{sqrt{2pi sigma^2}}e^{-frac{1}{2sigma^2}(y_i – mu)^2} \ &= dfrac{1}{(2pi sigma^2)^{n/2}}e^{-frac{1}{2sigma^2}sum_{i = 1}^n (y_i – bar{y})^2}e^{-frac{1}{2sigma^2}n(bar{y} – mu)^2} end{align}

From Theorem 1, it follows that where $$T(mathbf{Y}) = (bar{Y}, sum_{i = 1}^n (Y_i – bar{Y})^2)$$ is a sufficient statistic for $$(mu, sigma)$$.

Theorem 1 is presented as follows:

A statistic $$T(mathbf{Y})$$ is sufficient for $$theta$$ if, and only if, for all $$theta in theta$$

$$L(theta; mathbf{y}) = g(T(mathbf{y}), theta) times h(mathbf{y})$$

where the function $$g(cdot)$$ depends on $$theta$$ and the statistic $$T(mathbf{Y})$$, while the function $$h(cdot)$$ does not contain $$theta$$.

Theorem 1 implies that the likelihood $$L(theta; mathbf{y})$$ depends on the data only through $$T(mathbf{y})$$, $$T(mathbf{Y})$$ is a sufficient statistic for $$theta$$ and $$h(mathbf{y}) equiv 1$$.

For reference to another example, here is a Poisson example that I recently posted:

Let $$Y_1, dots, Y_n$$ be a i.i.d. $$text{Pois}(lambda)$$. Then

begin{align} L(lambda; mathbf{y} &= prod_{i = 1}^n e^{-lambda} dfrac{lambda^{y_i}}{y_i!} \ &= e^{-lambda n} dfrac{lambda^{sum_{i = 1}^n y_i}}{prod_{i = 1}^n y_i!} \ &= g(T(mathbf{y}), lambda) times h(mathbf{y}) end{align}

where $$T(mathbf{y}) = sum_{i = 1}^n y_i$$, $$g(T(mathbf{y}), lambda) = e^{-lambda n} lambda^{T(mathbf{y})}$$ and $$h(mathbf{y}) = dfrac{1}{prod_{i = 1}^n y_i!}$$

There are three things that I don’t understand here:

1. How is it that $$sum_{i = 1}^n (y_i – mu)^2 = sum_{i = 1}^n (y_i – bar{y})^2 + n(bar{y} – mu)^2$$?

2. If, for $$L(theta; mathbf{y})$$, we require the form $$g(T(mathbf{y}), theta) times h(mathbf{y})$$, then, for $$L(mu, sigma; mathbf{y})$$, what form do we require? Trying to think of this myself, I thought of three potentially correct forms: $$g(T(mathbf{y}), (mu, sigma)) times h(mathbf{y})$$, $$g(T(mathbf{y}), (sigma, mu)) times h(mathbf{y})$$, or $$g(T(mathbf{y}), mu, sigma) times h(mathbf{y})$$.

3. Related to 2., comparing the first example to the Poisson example, I don’t understand the conclusion of the first example. How does $$T(mathbf{Y}) = (bar{Y}, sum_{i = 1}^n (Y_i – bar{Y})^2)$$ satisfy the form $$g(T(mathbf{y}), lambda) times h(mathbf{y})$$?

I would greatly appreciate it if people would please take the time to clarify these points.

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