# Simplification of bivariate normal $\phi_2(x,y,\rho)$ at $y=y_F$ (i.e. fixing one of the axes)

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.

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:

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

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

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

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:

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

• Are you asking for the pdf of $X\mid Y$ where $(X,Y)$ is bivariate normal? – StubbornAtom Apr 18 at 7:00
• 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 $\phi_1$. – Felipe D. Apr 18 at 7:20
• 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. – Sanket Agrawal Apr 18 at 19:03
• 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. – Sanket Agrawal Apr 18 at 19:07
• You're right, @SanketAgrawal, I was getting my notation mixed up. I finally understand what the heck was going on. Thanks for the help! – Felipe D. Apr 18 at 23:03