I'm trying to solve the following maximum likelihood estimation but for multiplicative noise instead of additive noise:

So the goal is to do ML-estimation for a scalar constant $x$, which is multiplied with noise:

$$\tilde{d}[n] = a[n]x$$

The noise $a[n]$ is a realization of the i.i.d. random variable $A \sim \mathcal{N}(\mu_A, \sigma_A)$.

The probability density function is given by

$$ \pi(x,\mathbf{\tilde{d}}) = \frac{1}{x^N} \pi_A\left(\frac{\tilde{\mathbf{d}}}{x}\right)$$

with $\pi_A$ being the pdf of $A$.

Now I took the natural logarithm of this expression and derived with respect to $x$, which let to the following term:

$$\frac{\partial}{\partial x} \ln\left(\pi(x,\mathbf{\tilde{d}})\right) = -\frac{N}{x} + \frac{1}{\sigma_A^2} \sum_{n=0}^{N-1} \left[ \frac{\tilde{d}[n](\tilde{d}[n]-\mu_a x)}{x^3} \right] \overset{!}{=}0$$

But how do I continue from that? My idea was to solve this equation for $x$, but I can't figure out a closed form solution...

  • $\begingroup$ Hi: Assuming that $x$ is the parameter that you're trying to estimate (we usually don't denote them as $x$ which is why I'm confirming ), you can take the log of both sides so that you almost have OLS . But note that, if you do that, then you end up with noise that is lognormal so you need to maximize that likelihood rather than minimize the sum of the squared errors like you would with OLS. $\endgroup$
    – mark leeds
    Oct 12, 2020 at 15:49
  • $\begingroup$ @markleeds, I think you're wrong here in the use of log normal. The log normal distribution means it is distributed normally when applying the log. Here it is distributed normally before the log. $\endgroup$
    – Royi
    Oct 13, 2020 at 7:01
  • $\begingroup$ Royi: You are correct. I said the wrong thing about the noise being lognormal. Phinie: I'm not sure what the distribution of the resulting term is so best to use Florian's approach. Thanks to Royi. $\endgroup$
    – mark leeds
    Oct 13, 2020 at 12:40

1 Answer 1


It looks okay to me. If you define empirical mean $\hat{\mu}_d = \frac 1N \sum \tilde{d}[n]$ and empirical second moment $\hat{\gamma}_d = \frac 1N \sum \tilde{d}^2[n]$ then you effectively have an equation of the form $$-N/x + A/x^3 + B/x^2 \stackrel{!}{=} 0,$$ where $A$ and $B$ directly depend on $\hat{\mu}_d, \hat{\gamma}_d, \mu_A, \sigma_A$. If you multiply this equation with $x^3$ it becomes a quadratic equation in $x$, which you can solve for $x$...

I get something like $$x^2 - \frac{\hat{\mu}_d \mu_A}{\sigma_A^2} x + \frac{\gamma_d}{\sigma_A^2} = 0$$ but that was scribbled down in a rush so I would have to check it again later. Can you confirm or correct?

  • $\begingroup$ thank you! I tried that in matlab and got good results! $\endgroup$
    – Phobos
    Oct 13, 2020 at 8:17
  • $\begingroup$ Awesome, glad to hear it helped! :) $\endgroup$
    – Florian
    Oct 13, 2020 at 8:35

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