Suppose $(X,Y)$ has the bivariate normal distribution with mean $(1,0)$ and variance $(1,1)$ and correlation coefficient $.5$. What is the conditional distribution of $Y$ given $X=1$?
Now, I am brand new to finding conditional distributions. What is the most straight forward way to go about this?
I think that I presume I would have to use the correlation coefficient formula, but I am honestly stumped. I ended up getting a normal random variable with mean $1$ and variance $.75$
If $Z$ is standard normal and independent of $X$, then the quantity $$\rho\cdot \frac{X-\mu_X}{\sigma_X} + \sqrt{1-\rho^2} \cdot Z$$ is normal with mean $0$, variance $1$, and correlation $\rho$ with $X$. In particular, $$\left(X, \sigma_Y\left(\rho\cdot \frac{X-\mu_X}{\sigma_X} + \sqrt{1-\rho^2} \cdot Z\right) + \mu_Y\right)$$ is bivariate normal with mean $(\mu_X, \mu_Y)$, variance $(\sigma_X^2, \sigma_Y^2)$, and correlation $\rho$.
Because $X$ and $Z$ are independent, the conditional distribution of the second component given $X=1$ can be obtained by plugging in $X=1$ into $\sigma_Y\left(\rho\cdot \frac{X-\mu_X}{\sigma_X} + \sqrt{1-\rho^2} \cdot Z\right) + \mu_Y$. What remains is a linear function of $Z$, and you can show that it follows a normal distribution with mean $\rho\frac{\sigma_Y}{\sigma_X} (1-\mu_X) + \mu_Y$ and variance $\sigma_Y^2(1-\rho^2)$.