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When counting the score for poisson distribution I get the log likelihood

$$S(\mu ) = \frac{\partial \ell(\lambda )}{\partial \lambda } = \sum_1^n \left(\frac{y_i}{\lambda}-1\right)$$

Textbook says that it is equivalent to: $$\frac{n(\overline{y}-\lambda)}{\lambda}$$

I can get easily solve the fisher score from there on, but I'm not quite sure about this equation. Why does it switch to the mean of y? If you take the $1/{\lambda}$ on the front of the sum, wouldn't it then be $$S(\mu ) = \frac{\partial \ell(\lambda )}{\partial \lambda } = \left(\frac 1 \lambda \right) \sum_1^n (y_i-1) \text{ ?}$$

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    $\begingroup$ Two Greek letters appear here: $\mu$ and $\lambda$. Is that a typo? Did you mean $S(\lambda)$ rather than $S(\mu)\text{?} \qquad$ $\endgroup$ – Michael Hardy May 7 '19 at 17:27
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The summation notation is a linear operator. So expand the fraction:

$$\begin{eqnarray} \sum_{i=1}^n \left( y_i / \lambda - 1 \right) &=& \sum_{i=1}^n \left( (y_i - \lambda) / \lambda \right) \\ &=& \left( \left(\sum _{i=1}^ny_i - n \lambda \right) / \lambda \right) \\ &=& n \left( \left(\bar{y} - \lambda \right) / \lambda \right) \\ \end{eqnarray}$$

Note that the score takes the canonical form of (the sum of) sample mean minus the theoretical mean over the variance.

You will find a similar canonical form of the score function in most of the exponential families like binomial, gamma, normal, and so on.

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\begin{align} & \sum_{i=1}^n \left( \frac {y_i} \lambda -1\right) = \sum_{i=1}^n \frac {y_i-\lambda} \lambda \\[10pt] = {} & \frac{y_1-\lambda}\lambda + \cdots + \frac{y_n-\lambda} \lambda \\[10pt] = {} & \frac{(y_1 + \cdots + y_n) - n\lambda} \lambda = \frac{n\overline y - n\lambda} \lambda. \end{align}

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  • $\begingroup$ I still don't understand why does the $$({y_1} +...+{y_n})$$ switch to the mean? Isn't $$({y_1} +{y_2} + {y_3} + ...+{y_n})$$ just the sum of y's over 1 to n? $\endgroup$ – Guest192 May 7 '19 at 19:56
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    $\begingroup$ @Guest192 : The sum did not switch to $\overline y,$ but rather to $n\overline y.$ The sum is $n$ times the mean. $\qquad$ $\endgroup$ – Michael Hardy May 8 '19 at 16:46

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