Maximum likelihood function (MLE) for Levy distribution

Question

I am a student who is writing a little thesis belonged in the applied mathematics category.

I choose a "Levy distribution" defined as, \begin{equation} \lambda(t;u,c) = \begin{cases} \sqrt{\frac{c}{2\pi}}\frac{ e^{-\frac{c} {2(t-u)} }} {(t-u)^{3/2}} & \text{if } t>u \\ 0 & \text{otherwise} \end{cases} \end{equation} to examine the behavior of the variables.

Now I thought I had to apply the MLE to obtain the good estimation of the parameter: $c,u $

However, after calculating the derivative of logarithm of the levy distribution. I obtained only \begin{equation} \frac{d\lambda(t;u,c)} {dc} = \frac{c-t+u} {2ct-2cu} = 0 \iff c = t-u \end{equation} and \begin{equation} \frac{d\lambda(t;u,c)} {du} = -\frac{c-3t+3u} {2(t-u)^{2}} = 0 \iff u = t-\frac{c}{3} \end{equation}

However, both the parameters $c$ and $u$ are dependent on each other. Could somebody suggest me the better method for estimating?

Thank you very much.

Answer 1

Parameter $u$ in your Levy pdf is just a shift parameter that adjusts the lower bound from 0. It is equivalent to the transformation $X = T + u$, where random variable $T$ has pdf $f(t)$:

The following plot illustrates the pdf, as parameter $c$ changes:

From the plot, the lower bound parameter will generally be clearly defined by the data itself (unless parameter $c$ is very large) ... So, the way I would proceed would be to:

a) zero-shift your data so that the lower bound is 0; i.e. newdata = data - min[data]

b) then your problem reduces to finding the MLE of parameter $c$, for the single parameter model.

That can be solved by hand (if you prefer), or get a computer to do the work for you. Here I am using the mathStatica add-on to Mathematica (I am one of the authors of the former). mathStatica has a special SuperLog function specially designed for formulating log-likelihood problems:

Then, for a random sample of size $n$ drawn on $T$, the log-likelihood function is:

The score function is the gradient of the log-likelihood with respect to $c$:

Setting the score to zero and solving for $c$ corresponds to the first-order conditions:

Finally, check the second-order condition, evaluated at solc:

... reveals that the Hessian is strictly negative at the FOC.

Therefore, the MLE of $c$ is:

$$\hat{c}=\frac{n}{\underset{i=1}{\overset{n}{\sum }}\frac{1}{T_i}.}$$

Answer 2

Your post present a computation when the sample $\{t\}$ has size $1$. Then $\partial_c\log\lambda$ has the sign of $(t-u)-c$ and $\partial_u\log\lambda$ has the sign of $c-3(t-u)$ (watch out for the signs in the post...). The domain of $(c,t)$ is $c\gt0$ and $0\leqslant u\lt t$.

Let $\ell(c,v)=\lambda(t;c,t-v)$. Then, for every $(c,v)$, $$ \ell(c,v)\leqslant\ell(v,v)\leqslant\ell(v,\tfrac13v)\leqslant\ell(\tfrac13v,\tfrac13v)\leqslant\ell(\tfrac13v,\tfrac19v)\leqslant\ldots $$ This shows that the supremum of $\ell$ is reached when $(c,v)\to(0,0)$, hence the supremum of $\lambda(t;c,u)$ for some given $t$ is reached when $(c,u)\to(0,t)$. To check this, note that $$ 2\log\lambda(t;c,u)=\log c-\frac{c}{t-u}-3\log(t-u), $$ hence $\lambda(t;\varepsilon,t-\varepsilon)=-1-2\log\varepsilon\to+\infty$ when $\varepsilon\to0$. To sum up, using the MLE approach for a sample of size $1$ seems to be a bad idea. Some remarks are in order here.

• First, this degenerate behaviour probably disappears when the sample $\{t_1,t_2,\ldots,t_n\}$ has size $n\geqslant4$ (computations omitted).

• Second, if $u$ is known, optimizing $\lambda(t;c,u)$ over $c$ yields $\hat c=t-u$ when the sample has size $1$ and, for an i.i.d. sample of size $n$, $$ \frac{n}{\hat c}=\sum_{k=1}^n\frac1{t_k-u}. $$

• Third, if $c$ is known, optimizing $\lambda(t;c,u)$ over $u$ yields $\hat u=t-\frac13c$ when the sample has size $1$ and, for an i.i.d. sample of size $n$, $\hat u$ should solve $$ c\sum_{k=1}^n\frac1{(t_k-\hat u)^2}=3\sum_{k=1}^n\frac1{t_k-\hat u}. $$