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Suppose we start off with the traditional standard bivariate normal distribution:

$$\phi_2(x,y|\rho,\mu_x=0,\mu_y=0,\sigma_x=1,\sigma_y=1)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp \left(-\frac{x^2-2\rho x y + y^2}{2(1-\rho^2)}\right)$$

where the $\mu$s are the mean parameters, the $\sigma$s are the standard deviation parameters and $\rho$ is the correlation term between $x$ and $y$.

My question is: what happens to $\phi_2$ when we fix $y=y_F$? In other words, what happens when the only "free" variable in $\phi_2$ is $x$?

In this case, can we express $\phi_2$ in terms of a univariate normal distribution $\phi_1$?


Edit

After banging my head against my table for quite a bit, I noticed that I was confusing two different concepts:

  • a bivariate normal distribution with $y$ fixed at some point $y_F$
  • the conditional bivariate distribution with a given $y$ (i.e. $prob(X=x|Y=y_F)$)

These are two very different beasts, but I was treating them as the same.

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  • $\begingroup$ Are you asking for the pdf of $X\mid Y$ where $(X,Y)$ is bivariate normal? $\endgroup$ Apr 18 '19 at 7:00
  • $\begingroup$ Yes, exactly. I know we can express the pdf of $X|Y$ by the traditional conditional distribution rule $\phi_2(X|Y)=\frac{\phi_2(X,Y)}{\phi_1(Y)}$. But this solution still requires the calculation of a bivariate pdf $\phi_2$. I'm trying to simplify things such that, in the end, I only need to deal with univariate distributions $\phi_1$. $\endgroup$
    – Felipe D.
    Apr 18 '19 at 7:20
  • $\begingroup$ The ratio is obvious since it is equal to the inverse of the value of PDF of Y at Y=1 and as long as you keep Y=1, no matter what X you take, the ratio will remain the same. Moreover, in your comment you write $\phi_2(X|Y)$, which is wrong. Following your notations, since X|Y is a univariate random variable, it's distribution should be represented by $\phi_1(X|Y)$ instead. $\endgroup$ Apr 18 '19 at 19:03
  • $\begingroup$ A bivariate random variable (X,Y) and a random variable (X|Y) are two different random variables. I think you are confusing between the two. $\endgroup$ Apr 18 '19 at 19:07
  • $\begingroup$ You're right, @SanketAgrawal, I was getting my notation mixed up. I finally understand what the heck was going on. Thanks for the help! $\endgroup$
    – Felipe D.
    Apr 18 '19 at 23:03
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Here's how you can calculate $\phi_2(x,y,\rho)$ when $y$ is fixed at $y_F$.

First, from basic conditioning rules, we know that:

$$ \text{prob}(X=x|Y=y_F) = \frac{\text{prob}(X=x,Y=y_F)}{\text{prob}(Y=y_F)} \Rightarrow$$

$$ \text{prob}(X=x,Y=y_F) = \text{prob}(X=x|Y=y_F) \cdot \text{prob}(Y=y_F) \Rightarrow$$

$$ \phi_2(x,y_F,\rho) = \text{prob}(X=x|Y=y_F) \cdot \phi_1(y_F|\text{mean}=0,\text{var}=1) $$

Now,using the information from the conditional probability of bivariate normal distributions, we can swap out that $\text{prob}$ term in the middle by an actual function:

$$ \phi_2(x,y_F,\rho)=\phi_1(x|\text{mean}= \rho y_F, \text{var}=1-\rho^2) \cdot \phi_1(y_F|\text{mean}= 0, \text{var}=1)$$

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