# Maximum likelihood estimator for multiplicative Gaussian noise

So I'm trying to derive an analytical solution for a MLE that should estimate a static value polluted by multiplicative Gaussian noise.

The vector of measurements $\tilde{\boldsymbol{d}}$ is given as $a[n]x_u$ where $a[n]$ is the $n$-th realization of the random variable $A \propto \mathcal{N}(\mu_A, \sigma_A)$ and $x_u$ is an unknown constant that should be estimated from $\tilde{\boldsymbol{d}}$.

I got as far as the log-likelihood function, which I now need to maximize with regard to $x$. To do this, I need to take the derivative of the log-likelihood , set it to zero and solve for $x$. However, the summation term at the end is giving me a headache as I can't figure out its derivative.

$\ell(x, \tilde{\boldsymbol{d}}) = -N \cdot \log x - N\cdot \log\sigma_A - \frac{N}{2} \log 2\pi - \frac{1}{2\sigma^2_A}\sum\limits_{n=1}^N\bigg(\frac{\tilde{d}[n]}{x}-\mu_A\bigg)^2$

• Could you include the distribution of your "static value polluted by multiplicative Gaussian noise" ? Commented Jul 8, 2016 at 19:16
• @Gilles I have edited the question to include the original problem. Commented Jul 9, 2016 at 11:38
• Thanks ! I think you meant $A \sim \mathcal N (\mu_A, \sigma^2_a)$. Commented Jul 9, 2016 at 14:55
• Btw is $\tilde d$ given as $a[n] x_u$ with $x_u$ fixed or is given as $a[n] x$ with $x$ variable ? Commented Jul 9, 2016 at 15:21
• please see my answer below. Commented Jul 9, 2016 at 16:01

OK, let's have a look at one of the problematic terms: $$\frac{\delta}{\delta x} \bigg[ \bigg(\frac{\tilde{d}[n]}{x}-\mu_A\bigg)^2 \bigg ] = - \frac{2 \tilde{d}[n] \bigg (\tilde{d}[n] - \mu_A x\bigg) }{x^3}$$ which can be verified by Wolfram Alpha.

The full derivative of the summation term is then just this summed over $n$.

• Ah yes, of course all summation terms can be treated independently. Thanks! Commented Jul 9, 2016 at 11:50

To be clear:

I'm assuming $\mathbf{\tilde{d}}$ is given as $a[n]x$, with a deterministic variable $x$, so that with the i.i.d. on $\mathbf{\tilde{d}}$ you then have $$\text{if}\quad A \sim \mathcal (\mu_A, \sigma^2_A) \Longrightarrow \mathbf{\tilde{d}} \sim \mathcal (\mu_A x, \sigma^2_A x^2)$$ And with this, only the first and last term of your log-likelihood depend on $x$; I'm assuming your $\log(\cdot)$ is the natural logarithm $\ln (\cdot)$, otherwise you'll have to adjust the quadratic term accordingly. And this then gives you \begin{align} \frac{\textrm{d}\left[\ell(x, \tilde{\boldsymbol{d}})\right]}{\textrm{d}x}&=\frac{-N}{x}-\frac{1}{2\sigma^2_A}\sum_{n=1}^N\left[ -2\frac{\tilde{d}[n]}{x^2}\left(\frac{\tilde{d}[n]}{x}-\mu_A\right)\right]\\ &=\frac{-N}{x}+\frac{1}{\sigma^2_A{x^3}}\sum_{n=1}^N\tilde{d}[n]\left(\tilde{d}[n]-\mu_A x\right) \end{align} You can proceed from here.

• Why do you say that all x in the log-likelihood would become $x_u$? This does not make sense to me. $x_u$ is only contained in $\tilde{\boldsymbol{d}}$. Also $x$ is not a function of $n$. If our estimator is unbiased, then at the maximum of $\ell$ with regard to $x$, $x$ is the estimate of $x_u$. Commented Jul 9, 2016 at 17:20
• @sobek you're right, $x$ is not a function of $n$. And wrong formulation. However, if $x_u$ is to be estimated then your PDF is a function $x$ (and not $x_u$), $\mu_A$, and $\sigma^2_A$. Commented Jul 9, 2016 at 19:59
• The PDF of the estimator, yes, but the PDF of $\tilde{\boldsymbol{d}}$, this i don't really understand. Thanks for the hint about the logarithm, you are right, since it is only used to make it easier to deal with the exponential of the Gaussian PDF, the natural logarithm must be used. Commented Jul 9, 2016 at 20:35