1. ## Bivariate normal distribution

Assume that $(X, Y)$ is a bivariate normal with $\mu_X = 1$ and $\sigma_X = .5$ and I want to find $P(X \leq .5)$. Is it true that $f_X \sim N(\mu_X, \sigma_X$)? Then I can just integrate that from $-\infty$ to .5. Is this correct?

2. Looks good to me.

3. Using that method, I get

$f_X = \frac{1}{.5\sqrt{2\pi}}e^{-2(x-1)^{2}}$

To find $P(X \leq .5)$, I take the integral.

$\int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-2(x-1)^{2}}dx$

How do I do that integral?

4. You don't, it's intractable (when I said "looks good" I didn't think you would try to integrate analytically) . Standardize X and use a Z-Table or software. The question is essentially asking the probability of X being one standard deviation below the mean, or more; should be, what, 16% or so?

5. Now I want to find $P(X \leq 0 | Y = -.2)$
Do I find $E(X|Y = -.2$) and $Var(X|Y = -.2)$ and use that in the standard normal?

6. Originally Posted by BERRY
Now I want to find $P(X \leq 0 | Y = -.2)$
Do I find $E(X|Y = -.2$) and $Var(X|Y = -.2)$ and use that in the standard normal?
$X|Y = y$ is also distributed normally. It should be a result from your text that $(Y|X = x) \sim N(\mu_y + \rho (\sigma_y / \sigma_X)(x - \mu_X), \sigma^2 _Y (1 - \rho^2))$.

7. ## It's incorrect...

Originally Posted by BERRY
Is it true that $f_X \sim N(\mu_X, \sigma_X$)?
Corrected: $f_X \sim N(\mu_X, \sigma_X^2$)

That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.

8. Originally Posted by paolopiace
Corrected: $f_X \sim N(\mu_X, \sigma_X^2$)

That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.
What if Cov(X,Y) isn't 0? I was given that $\rho = .6$

9. Originally Posted by BERRY
What if Cov(X,Y) isn't 0? I was given that $\rho = .6$
rho = Covariance(X,Y) / (sigma_X * sigma_Y)

I suggest that you post all the data of the problem. What is sigma_Y ?

10. Originally Posted by paolopiace
Corrected: $f_X \sim N(\mu_X, \sigma_X^2$)

That assumes Cov(X,Y) = 0 which does not imply independence. On the other end, independence implies Cov(X,Y) = 0.
The marginal distribution of X does not depend on the Covariance. The conditional distribution does. Moreover, for the Bivariate Normal, Cov(X, Y) = 0, DOES imply independence.

11. Originally Posted by BERRY

To find $P(X \leq .5)$, I take the integral.

$\int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-2(x-1)^{2}}dx$

How do I do that integral?
I am sure you meant:

$\int_{-\infty}^{.5} f_X dx = \frac{1}{.5\sqrt{2\pi}}\int_{-\infty}^{.5}e^{-\frac{1}{2}(\frac{x-1}{.5})^2}dx$

12. Those are the same