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I am given the following PDF:

$$f_X(z) = \begin{cases} \frac{ak^a}{z^{a+1}}, & \text{for $z \ge k$} \\ 0, & \text{for $z \lt k$} \end{cases}$$

For some parameters $k$ and $a$. I am then told to find the Maximum-Likelihood estimator for $a$ if we know $k$ and have $n$ independent observations.

The Likelihood function is $$L_z(a,k) = \prod_{i=1}^{n}\frac{ak^a}{z_i^{a+1}}$$ and then the Log-Likelihood function is $$l_z(a,k) = log(L_z(a,k)) = \sum_{i=1}^{n}log(\frac{ak^a}{z_i^{a+1}}) = n \cdot \Bigl( a \cdot log(k) + log(a) \Bigr)- (a+1) \sum_{i=1}^{n}log(z_i)$$

Now we take the derivative with respect to $a$ and set it equal to $0$:

$$n \cdot (log(k) + \frac{1}{a}) - \sum_{i=1}^{n}log(z_i) = 0$$ $$\iff \frac{1}{a} = \frac{1}{n}\sum_{i=1}^{n}[log(z_i)] - log(k)$$ $$\iff a = \Bigl(\frac{1}{n}\sum_{i=1}^{n}[log(z_i)] - log(k)\Bigr)^{-1}$$

However, this estimator depends on the base of the logarithm, which shouldn't be possible. Where did I go wrong?

yagod
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1 Answers1

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In your step where you differentiated with respect to $a$, you used $\frac{d}{da}(\log a) = \frac{1}{a}$; in doing this, you assumed a log base of $e$.

If you redo the calculations with a general log base $b$ (where $\log_b z =\frac{\log z}{\log b}$ and $\frac{d}{dz}(\log_b z)=\frac{1}{(\log b)z}$ and $\log$ is the log with base $e$), you will see you end up at the same result, because a $\frac{1}{\log b}$ factor cancels when setting the derivative to $0$.

Minus One-Twelfth
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