# Why is this likelihood function equal to the noise PDF?

My professor has this slide up 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...

• Where does it say p(w)=Likelihoodfunction? The function p(y|f) is a conditional probability density. Jun 25 '14 at 17:38
• @emcor What do you mean? p(w) = p(y|f) as seen above, and p(y|f) is the likelihood function as given here Jun 25 '14 at 17:41
• 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.
– whuber
Jun 25 '14 at 17:55
• 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 $f$.
– whuber
Jun 25 '14 at 18:10
• 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.
– whuber
Jun 25 '14 at 18:44

## 1 Answer

$p(w)=p(y|f)$:

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

$w=y-Hf$

,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:

$p(w)=p(y-Hf|f,H)=p(y|f)$.

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 $w$.

• 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.
– whuber
Jun 25 '14 at 18:24
• 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... Jun 25 '14 at 18:26
• "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. Jun 25 '14 at 18:32
• 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.
– whuber
Jun 25 '14 at 18:35
• @whuber Thanks whuber, let me study that link and digest. Jun 25 '14 at 19:55