3
$\begingroup$

I've been reading a text containing an introduction to probability theory, and I ran into the following formula for conditional probability distributions. Note that $Pr(x)$ here is not the probability of an event, but the PDF of a random variable $x$.

$$\operatorname{Pr}(x\mid y=y^*)=\frac{\operatorname{Pr}(x,y=y^*)}{\int\operatorname{Pr}(x,y=y^*)dx}=\frac{\operatorname{Pr}(x,y=y^*)}{\operatorname{Pr}(y=y^*)}\tag{2.3},$$

The equation itself makes simple sense to me. We're just normalizing the joint PDF of $x$ and $y$ where $y = y^*$. The notation does seem a tad odd to me, though - I'm interpreting the numerator of the second expression as $\operatorname{Pr}(x, y) |_{y=y^*}$. Would that be accurate?

If so, then the second part is simple as well. We're just marginalizing the denominator to get a simpler expression. The denominator can then be expressed as $\operatorname{Pr}(y)|_{y=y^*}$, just the probability that $y=y^*$. My confusion comes in when the notation is simplified, though:

$$\operatorname{Pr}(x\mid y)=\frac{\operatorname{Pr}(x,y)}{\operatorname{Pr}(y)}.\tag{2.4}$$

The simplified notation looks nice, but we've already defined $\operatorname{Pr}(x, y)$ and $\operatorname{Pr}(y)$! In this sense, we're dividing a joint PDF by a single-variable PDF, which hasn't been defined, as far as I can tell. I believe I'm supposed to implicitly assume $y=y^*$ here, but that seems awfully arbitrary to me. It especially seems unintuitive to assume different variables when more variables become involved, such as in Bayes' Theorem:

$$\begin{align} \operatorname{Pr}(y\mid x) & = \frac{\operatorname{Pr}(x\mid y)\operatorname{Pr}(y)}{\operatorname{Pr}(x)}\\ & = \frac{\operatorname{Pr}(x\mid y)\operatorname{Pr}(y)}{\int\operatorname{Pr}(x,y)\,dy}\\ & = \frac{\operatorname{Pr}(x\mid y)\operatorname{Pr}(y)}{\int\operatorname{Pr}(x\mid y)\operatorname{Pr}(y)\,dy},\tag{2.9} \end{align}$$

In such a circumstance, am I supposed to see such a formula, associate $\operatorname{Pr}(x)$ with the $\operatorname{Pr}(y\mid x)$ on the other side, then assume $\operatorname{Pr}(x)$ is not the general distribution itself but $\operatorname{Pr}(x)$ at some $x=x^*$? Such a syntax appears rather imprecise and seems to involve a lot of guesswork - am I interpreting it correctly?

$\endgroup$
  • $\begingroup$ Thank you for the edit @IWantToRemainAnonymous! $\endgroup$ – Chandler Watson Sep 5 '15 at 16:30
  • $\begingroup$ Also thanks to @Math1000 for the \mid fix! $\endgroup$ – Chandler Watson Sep 5 '15 at 18:16
1
$\begingroup$

Isn't it funny what a good night's sleep can do to some faulty intuition?

Responding to the issue of dividing a joint PDF by a single-variable PDF, both are simply scalars, so we can just divide them pointwise like any other function. Equation 2.3 is true at any value $y=y^*$, so by unbinding the value of $y$ and allowing it to vary, we prove it for the entire distribution.

As such, it's not a notational issue. Both sides of equation 2.4 are exactly equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.