My professor has this slide up here:

enter image description here

Here, $y$ is an observed signal. $H$ is a deterministic transformation, which is assumed known. $f$ is the original signal (which we dont know), and $w$ is random gaussian noise. We are trying to recover $f$.

I understand everything, except for, why $p(\mathbf{w})$ = $p(\mathbf{y}|\mathbf{f})$.

That is, I understand that the multidimensional noise PDF is given by the above expression.

But why is that expression, ALSO equal to the likelihood function, $\mathbf{y}$, given $\mathbf{f}$? I'm not seeing this...

  • $\begingroup$ Where does it say p(w)=Likelihoodfunction? The function p(y|f) is a conditional probability density. $\endgroup$
    – emcor
    Jun 25, 2014 at 17:38
  • $\begingroup$ @emcor What do you mean? p(w) = p(y|f) as seen above, and p(y|f) is the likelihood function as given here $\endgroup$
    – Creatron
    Jun 25, 2014 at 17:41
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    $\begingroup$ The likelihood is a function of the parameters,, so notation like "$p(w)$" clearly does not refer to a likelihood. Unless a distribution is assumed for $f$, though, "$p(y|f)$" is not a conditional probability density, either: it merely refers to the probability density of $y$ as it depends on the parameters $f$. By assuming $W=Y-Hf$ has a Gaussian distribution, all you have to do is plug $y-Hf$ into the formula for a (multivariate) Gaussian density. Fixing $y$, $H$, and $C_{ww}$, it becomes a function of $f$: in that sense it's a likelihood. $\endgroup$
    – whuber
    Jun 25, 2014 at 17:55
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    $\begingroup$ We have to do some careful interpreting because the notation is sloppy. Apparently the model is $Y=Hf+W$ where $W$ is a random vector-valued variable. This makes $Y$ a random variable, too. Given any value of $f$, any realization of $Y$, which is written $y$, corresponds to a realization $y-Hf$ of $W$. The probability density of that realization is given by the equation. The right hand side $\Lambda$ is a function of $(y,H,C_{ww},f)$. If you assume values for $H$ and $C_{ww}$, and are given the data $y$, $f$ remains the only variable and you can study how $\Lambda$ depends on $f$. $\endgroup$
    – whuber
    Jun 25, 2014 at 18:10
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    $\begingroup$ Your edits support my suppositions about how to read the slide. The $f$ you are trying to recover plays the role of unknown parameters; everything else is either known or assumed. Thus the likelihood will be considered a function of $f$ and you will later find values of $f$ that make the likelihood as large as possible. You might go even further and deduce confidence limits for your estimates of $f$ by studying how the likelihood varies as you vary $f$ around its maximizing value. You might possibly even adopt a "prior distribution" for $f$, but that would not alter the present interpretation. $\endgroup$
    – whuber
    Jun 25, 2014 at 18:44

1 Answer 1



It is because of what is annotated in red on the slide, you have $w$ and $y$ linked as:


,so $p(w)=p(y-Hf)$ as well.

If $H$ and $f$ are held constant, $y$ is the only random variable which determines the probability:


I assume he omits $H$ because it is defined as constant anyways, so the probability is no longer dependent on $f$ neither on $H$.

He then correctly substitutes $'w=y-Hf'$ into the Gaussian density of $w$.

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    $\begingroup$ One point worth addressing is the potential confusion of data with random variables. If the right hand side is called a "likelihood," then "$y$" must refer to data (that is, a realization), not to a random variable. Furthermore, if we accept that calling the RHS a likelihood was intentional, then we must emphasize its dependence on $f$ rather than dismiss it. $\endgroup$
    – whuber
    Jun 25, 2014 at 18:24
  • $\begingroup$ Thank you Emcor. FYI, I edited my question to give more details on the background setup. That said, unfortunately I still find myself somewhat puzzled by why, exactly, p(w) = p(y|f). Specifically, I am not sure why you are saying that f is held constant, when we are in fact trying to find it... $\endgroup$
    – Creatron
    Jun 25, 2014 at 18:26
  • $\begingroup$ "Likelihood" is the density evaluated at a datapoint, which we have in p(w) as already clear. As we also have p(w)=p(y-Hf), it is a notational convention to write the parameter to be maximized in the Maximum Likelihood Function as $f(x|\theta)$. This convention might be confusing here. $\endgroup$
    – emcor
    Jun 25, 2014 at 18:32
  • $\begingroup$ Creatron, I believe you may be overthinking this. In a formula like $2-x^2$, $x$ has some definite but unknown value. Your circumstance is no different conceptually. You will estimate $f$ by maximizing the likelihood, just as you would consider varying the unknown $x$ to maximize $2-x^2$, even though the quantity $x$ refers to is whatever it is and that doesn't vary at all. For more about how the likelihood works I will refer you once more to the link I gave in an earlier comment: please read the thread at stats.stackexchange.com/questions/2641. $\endgroup$
    – whuber
    Jun 25, 2014 at 18:35
  • $\begingroup$ @whuber Thanks whuber, let me study that link and digest. $\endgroup$
    – Creatron
    Jun 25, 2014 at 19:55

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