# Deriving the Maximum Likelihood Estimation (MLE) of a parameter for an Inverse Gaussian Distribution

Given the following likelihood function

$$f(y|x,\tau) = \prod_{i=0}^Nf_T(u_i-x_i-\tau) \tag{1}$$

where, $$f_T(t)$$ is the probability density function of an Inverse Gaussian distribution given by

$$f_T(t) = \sqrt\frac{\lambda}{2\pi t^3} \exp\Bigl(- \frac{\lambda (t-\mu)^2}{2\mu^2t}\Bigr)\tag{2}$$

The goal here is to determine the MLE of parameter $$\tau$$

$$\hat{\tau}_{MLE} := \mathop{argmax}\limits_\tau f(y|x,\tau) \tag{3}$$

According to the principle of MLE and substituting $$(2)$$ in $$(1)$$, we will obtain the folowing

\begin{align}L(\tau) & = \prod_{i=1}^N \sqrt\frac{\lambda}{2\pi (u_i-x_i-\tau)^3} \exp\Bigl(- \frac{\lambda (u_i-x_i-\tau-\mu)^2}{2\mu^2(u_i-x_i-\tau)}\Bigr) \\\\ & =\Bigl(\frac{\lambda}{2\pi }\Bigr)^{N/2} \prod_{i=1}^N(u_i-x_i-\tau)^{-3/2} \exp\Bigl(- \frac{\lambda }{2\mu^2} \sum_{i=1}^N \frac{(u_i-x_i-\tau-\mu)^2}{u_i-x_i-\tau}\Bigr) \tag{4}\end{align}

Taking log, we get

\begin{align} logL(\tau) & = \frac{N}{2} log \Bigl(\frac{\lambda}{2\pi }\Bigr) - \frac{3}{2}\sum_{i=1}^N \log (u_i-x_i-\tau) - \frac{\lambda }{2\mu^2} \sum_{i=1}^N \frac{(u_i-x_i-\tau-\mu)^2}{u_i-x_i-\tau} \tag{5}\end{align}

Now taking the deriative w.r.t. $$\tau$$

\begin{align} \frac{d(logL(\tau))}{d\tau}& = 0 - \frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}(-1) - \frac{\lambda }{2\mu^2} \sum_{i=1}^N \left(\frac{2(u_i-x_i-\tau-\mu)}{u_i-x_i-\tau}(-1) - \frac{(u_i-x_i-\tau-\mu)^2}{(u_i-x_i-\tau)^2}(-1) \right)\\\\ & =\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda }{2\mu^2} \sum_{i=1}^N \left(\frac{-2(u_i-x_i-\tau-\mu)}{u_i-x_i-\tau} + \frac{(u_i-x_i-\tau-\mu)^2}{(u_i-x_i-\tau)^2}\right) \tag{6} \end{align}

Setting equation $$6$$ to $$0$$

\begin{align}\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda }{2\mu^2} \sum_{i=1}^N \left(\frac{-2(u_i-x_i-\tau-\mu)}{u_i-x_i-\tau} + \frac{(u_i-x_i-\tau-\mu)^2}{(u_i-x_i-\tau)^2} \right)= 0 \tag{7}\end{align}

Before getting to the problem in hand, is the derivations performed so far correct ?

Here is the bottleneck:

How do I proceed from here? The second summation term has become very complicated and I can't figure out how to derive $$\tau$$.

[UPDATE 2] as per the inputs from @gunes

$$=>\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda }{2\mu^2} \sum_{i=1}^N \left(-1+1^2 - 2*1*\left(\frac{u_i-x_i-\tau-\mu}{u_i-x_i-\tau}\right) + \left(\frac{u_i-x_i-\tau-\mu}{u_i-x_i-\tau} \right)^2\right)= 0$$

$$=>\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda }{2\mu^2} \sum_{i=1}^N -1+\left(\frac{\require{cancel} \cancel{u_i}-\require{cancel} \cancel{x_i}-\require{cancel} \cancel{\tau} -\require{cancel} \cancel{u_i}+\require{cancel} \cancel{x_i}+\require{cancel} \cancel{\tau}+\mu}{u_i-x_i-\tau} \right)^2= 0$$

$$=>\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda }{2\mu^2} \sum_{i=1}^N -1+\left(\frac{\mu}{u_i-x_i-\tau} \right)^2= 0$$

$$=>\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}+\frac{N\lambda}{2\mu^2 }- \frac{\lambda N\require{cancel} \cancel{\mu^2} }{2\require{cancel} \cancel{\mu^2}} \sum_{i=1}^N \frac{1}{(u_i-x_i-\tau)^2} = 0$$

$$=>\frac{3}{2}\sum_{i=1}^N \frac{1} {(u_i-x_i-\tau)}- \frac{\lambda N }{2} \sum_{i=1}^N \frac{1}{(u_i-x_i-\tau)^2}= -\frac{N\lambda}{2\mu^2 }$$

$$=>\sum_{i=1}^N \frac{3(u_i-x_i-\tau) - \lambda}{(u_i-x_i-\tau)^2} = -\frac{N\lambda}{\mu^2 }$$

$$=>\sum_{i=1}^N \frac{ \lambda-3(u_i-x_i-\tau)}{(u_i-x_i-\tau)^2} = \frac{N\lambda}{\mu^2 }$$

[UPDATE 3] As per the derivation provided by @Ben

$$1 + 3 H_{-1}(\tau)^2 H_1(\tau)^2 - 5 H_{-1}(\tau) H_1(\tau) + H_1(\tau)^2 H_{-2}(\tau) = 0.$$

As @Ben states below, we are left with the aforementioned equation which is obviously not straight forward to estimate $$\tau$$.

We are now left with the following questions: How can we solve this numerically? Are there sofware packages that can perform such kind of numerical solutions? Or is it better to write one ourselves?

• Why not cross multiply the two summands and then expand out. You can probably cancel off a fair few terms. PS. That sum includes both of the last two terms, so you should really include brackets to make that clear
– user220802
Sep 23, 2018 at 6:28
• @Xiaomi I have done something similar to that but could not arrive at a conclusive solution. I have added the sums in the brackets :) Sep 23, 2018 at 6:39
• In the first step of differentiation, $-2$ will be $2$; and in the line after you need the second term in the second summation positive. Therefore, in equation (7), you'll just change the signs of the first and second terms in the second summation. This expression can be made a square. Finally, I had $$\sum_{i=1}^{N}{\frac{\lambda-3(u_i-x_i-\tau)}{(u_i-x_i-\tau)^2}}=\frac{\lambda N}{\mu^2}$$ and stuck. Then, I tried it with $N=1$, and a quadratic equation emerged, requiring $9-\frac{4\lambda^2}{\mu^2}\geq 0$ for real solution. So, I don't have tips afterwards. Sep 23, 2018 at 14:41
• Your algebra doesn't make sense. (I'm afraid you may have been misled by some comments.) To see what's going on, consider the case $N=2$ and suppose $\lambda/(2\mu^2)=1.$ Write, say, "$y_i$" in place of $u_i-x_i.$ Just this simplification of notation might help clear things up.
– whuber
Sep 23, 2018 at 17:58
• Re your last edit: abstractly, you are solving an equation of the form $f(\tau)=0.$ This is known as univariate root finding. It's a basic tool of all software that performs any kind of numerical optimization and therefore is well discussed in books on numerical analysis, such as Numerical Recipes Because even this circumstance has various pitfalls, as in most scientific and numerical programming, it is best not to write your own unless you are very well versed in the theory already.
– whuber
Oct 1, 2018 at 16:41

The full derivation of the MLEs for IID data from an inverse Gaussian distribution can be found in the answer to this related question. In your case you have added an additional layer of complication by having observable data values $$t_i = u_i - x_i - \tau$$ that depend on some conditioning covariates and an additional parameter. From this formulation, your sampling density is:

$$f(\mathbf{u} | \mathbf{x}, \tau, \mu, \lambda) = \prod_{i=1}^n \Big( \frac{\lambda}{2 \pi (u_i-x_i-\tau)^3} \Big)^{1/2} \exp \Big( - \sum_{i=1}^n \frac{\lambda (u_i-x_i-\tau - \mu)^2}{2 \mu^2 (u_i-x_i-\tau)} \Big)$$

over the support $$\mathbf{u} \geqslant \mathbf{x} + \tau \mathbf{1}$$. The log-likelihood function is defined over $$\tau \leqslant \min (u_i-x_i)$$ and is given over this range by:

$$\ell_{\mathbf{u},\mathbf{x}}(\tau, \mu, \lambda) = \text{const} + \frac{n}{2} \ln (\lambda) - \frac{3}{2} \sum_{i=1}^n \ln (u_i-x_i-\tau) - \frac{\lambda}{2 \mu^2 } \sum_{i=1}^n \frac{(u_i-x_i-\tau - \mu)^2}{(u_i-x_i-\tau)}.$$

Finding the MLE: To facilitate our analysis we define the functions:

$$H_k(\tau) \equiv \frac{1}{n} \sum_{i=1}^n (u_i-x_i-\tau)^k.$$

We then have:

\begin{aligned} \frac{\partial \ell_{\mathbf{u},\mathbf{x}}}{\partial \tau}(\tau, \mu, \lambda) &= \frac{3}{2} \sum_{i=1}^n \frac{1}{u_i-x_i-\tau} + \frac{\lambda}{2 \mu^2 } \sum_{i=1}^n \frac{(u_i - x_i - \tau + \mu)(u_i-x_i-\tau - \mu)}{(u_i-x_i-\tau)^2} \\[10pt] &= \frac{3}{2} \sum_{i=1}^n \frac{1}{u_i-x_i-\tau} + \frac{\lambda}{2 \mu^2 } \sum_{i=1}^n \frac{(u_i - x_i - \tau)^2 -2 \mu (u_i-x_i-\tau) + \mu^2}{(u_i-x_i-\tau)^2} \\[10pt] &= \frac{3}{2} \sum_{i=1}^n \frac{1}{u_i-x_i-\tau} + \frac{\lambda}{2 \mu^2 } \Big[ n - 2\mu \sum_{i=1}^n \frac{1}{u_i-x_i-\tau} + \mu^2 \sum_{i=1}^n \frac{1}{(u_i-x_i-\tau)^2} \Big] \\[10pt] &= \frac{3n}{2} H_{-1}(\tau) + \frac{n \lambda}{2 \mu^2 } \Big[ 1 - 2 \mu H_{-1}(\tau) + \mu^2 H_{-2}(\tau) \Big]. \\[10pt] \end{aligned}

Taking $$\tau$$ to be fixed for the moment, the MLEs of the inverse Gaussian distribution are:

$$\hat{\mu}(\tau) = H_1(\tau) \quad \quad \quad \frac{1}{\hat{\lambda}(\tau)} = H_{-1}(\tau) - \frac{1}{H_1(\tau)}.$$

Substituting these functions yields:

\begin{aligned} \frac{\partial \ell_{\mathbf{u},\mathbf{x}}}{\partial \tau}(\tau, \hat{\mu}(\tau), \hat{\lambda}(\tau)) &= \frac{3n}{2} H_{-1}(\tau) + \frac{n}{2 H_1(\tau)^2 } \frac{1 - 2 H_1(\tau) H_{-1}(\tau) + H_1(\tau)^2 H_{-2}(\tau)}{H_{-1}(\tau) - H_1(\tau)^{-1}} \\[10pt] &= \frac{n}{2} \cdot \frac{1}{H_1(\tau)^2} \Big[ 3 H_{-1}(\tau) H_1(\tau)^2 - \frac{2 H_1(\tau) H_{-1}(\tau) - H_1(\tau)^2 H_{-2}(\tau) - 1}{H_{-1}(\tau) - H_1(\tau)^{-1}} \Big]. \\[10pt] \end{aligned}

Setting this partial derivative to zero yields the critical point equation:

$$1 + 3 H_{-1}(\tau)^2 H_1(\tau)^2 - 5 H_{-1}(\tau) H_1(\tau) + H_1(\tau)^2 H_{-2}(\tau) = 0.$$

This critical point equation will need to be solved numerically, as there is no simple expression for the solution.

• Thank you very much Ben. Although, I still have a question: How do I or rather where do I pick up to solve the problem numerically? For instance If I want to run it on a software like MatLab. Sep 26, 2018 at 7:48
• @nashynash: There are a number of numerical methods to find local maxima and minima in cases where these are not defined in closed form. Common methods include gradient descent and Newton-Raphson iteration. These can be done in some packages (not sure how for MatLab) or you can program them manually if you have familiarity with the mathematics. If you would like to solve the above numerically, perhaps post that as a new question?
– Ben
Sep 26, 2018 at 23:08
• @nashynash: Actually, on second thoughts, I see that you already have a bounty on this one, so maybe just leave your numerical question here.
– Ben
Sep 26, 2018 at 23:39
• Okay, Ben. I will do that. Sep 27, 2018 at 0:48