conditional normal distribution

michael27

New Member
dear all,

i have the following question that i don't get my head around.

suppose x1 and x2 are jointly normally distributed. i am interested in
the conditional distribution of x1|x2>c where c is a constant. i know
the following:

- E(x1|x2>c) and Var(x1|x2>c), which are functions of the inverse
Mills Ratio of the normal distribution. this looks like the problem is
solved, but my question is not to find out the conditional expectation
or variance but whether the pdf of x1|x2>c is normal?
- i also know E(x1|x1>c) and Var(x1|x1>c) and that the pdf of the
"cut-off normal" x1|x1>c is clearly not normal.
- finally i know the conditional distribution x1|x2 is normal and that
x1|x2>-infinity is normal as it is the marginal distribution of x1

do you have any idea or reference where to look this up?

thanks,
michael27

BGM

TS Contributor
Let $$(X_1, X_2) \sim \mathcal{N}_2(\mu_1, \mu_2, \sigma^2_1, \sigma^2_2, \rho)$$

The CDF of $$X_1|X_2 > c$$ is

$$\Pr\{X_1 \leq x|X_2 > c\} = \frac {\Pr\{X_1 \leq x, X_2 > c\}} {\Pr\{X_2 > c\}}$$

The denominator is just $$1 - \Phi\left(\frac {c - \mu_2} {\sigma_2}\right)$$, where $$\Phi$$ is the standard normal CDF.

The numerator is $$\int_{-\infty}^x \int_c^{+\infty} f_{X_1, X_2}(u,v)dvdu$$

Differentiating the numerator with respect to $$x$$, we have

$$\int_c^{+\infty} f_{X_1, X_2}(x,v)dv$$

$$= \int_c^{+\infty} \frac {1} {2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}$$$$\exp\left\{-\frac {1} {2(1-\rho^2)}\left[\frac {(x-\mu_1)^2} {\sigma_1^2} + \frac {(v-\mu_2)^2} {\sigma_2^2} - \frac {2\rho(x-\mu_1)(v-\mu_2)} {\sigma_1\sigma_2} \right]\right\}dv$$

$$= \frac {1} {2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\left\{-\frac {1} {2(1-\rho^2)}\left[\frac {(x-\mu_1)^2} {\sigma_1^2} - \frac {\rho^2(x-\mu_1)^2} {\sigma_1^2} \right]\right\}$$
$$\times \int_c^{+\infty} \exp\left\{-\frac {1} {2(1-\rho^2)} \left[\frac {v - \mu_2} {\sigma_2} - \frac {\rho(x - \mu_1)} {\sigma_1} \right]^2\right\}dv$$

$$= \frac {1} {2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\left\{-\frac {1} {2(1-\rho^2)}\left[\frac {(x-\mu_1)^2} {\sigma_1^2} - \frac {\rho^2(x-\mu_1)^2} {\sigma_1^2} \right]\right\}$$
$$\times \sqrt{2\pi\sigma_2^2(1-\rho^2)} \left[1 - \Phi\left(\frac {1} {\sigma_2\sqrt{1-\rho^2}}\left(c - \mu_2 - \frac {\rho\sigma_2(x-\mu_1)} {\sigma_1} \right)\right)\right]$$

$$= \frac {1} {\sqrt{2\pi\sigma_1^2}} \exp\left\{-\frac {1} {2} \frac {(x-\mu_1)^2} {\sigma_1^2} \right\} \times \left[1 - \Phi\left(\frac {1} {\sigma_2\sqrt{1-\rho^2}}\left(c - \mu_2 - \frac {\rho\sigma_2(x-\mu_1)} {\sigma_1} \right)\right)\right]$$

Combining you get the pdf of the conditional distribution. Sorry if there is any typo/mistakes