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Let

$$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim\mathcal{N}\left[\begin{pmatrix} 0\\ 0 \end{pmatrix} ,\begin{pmatrix} \sigma_{1}^2 & \sigma_{12} \\ \sigma_{12} & \sigma_{2}^2 \end{pmatrix} \right] $$

I know how to get the expression (in terms of normal pdf and cdfs) of the densities:

$$ f(x_1 | x_1 > \alpha) $$ (which is the truncated normal)

and

$$ f(x_1 | x_2 = \alpha) $$ (which is a conditional normal)

However, I'm stuck trying to figure out how to write

$$ f(x_1 | x_2 > \alpha) $$

in terms of normal pdfs and cdfs. It seems to me it should be a combination of the previous two steps, but I can't quite do it.

I tried doing this (Is this right?? I'm probably missing some normalization term)

$$ f(x_1 | x_2 > \alpha) = \int_\alpha^\infty g(x_1 | x_2 = u) \; f_2 (u) \; du $$

where $f_2 (\alpha)$ is the marginal distribution for $x_2$. Since I know the expression $ g(x_1 | x_2 = \alpha)$, I thought I could do this integral to get an expression in terms of a normal CDF but I haven't succeeded.

Please tell me if this is the way to go or I should try something simpler!!

Just to be crystal clear, the end objective is to get an expression for $ f(x_1 | x_2 > \alpha) $ in terms of normal PDFs and CDFs.

Thanks for your time.

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2 Answers 2

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You need a normalization factor for the integral. Basically the right hand side is the weighted sum of some conditional densities (Gaussians with possibly different mean and variance) then the normalization should be the sum of weighing factors. Namely it should read:

$$ f(x_1 | x_2 > \alpha) =\frac{\int_\alpha^\infty g(x_1 | x_2 = u) \; f_2 (x_2=u)\mbox{d}u}{\int_\alpha^\infty f_2 (x_2=u)\mbox{d}u} \, $$

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  • $\begingroup$ Thanks @seyhmus-gungoren ! Can you comment on whether I'm going in the right direction with this (basically just calculating the integral) or is there something easier I could do? $\endgroup$
    – cd98
    Mar 5, 2014 at 1:00
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I took an easier approach and I think I got it. Just to be a bit more general I'll be using $(\mu_1, \mu_2 ) $ as the means of $(\epsilon_1, \epsilon_2) $ even though the question stated they were zero.

Write the standard normal PDF and CDF by $\phi$ and $\Phi$.

Define $$Pr(\epsilon_2 > \alpha) = 1 - F_2 (\alpha) $$ and multiply and divide by this term:

$$ f (\epsilon_1 | \epsilon_2 > \alpha) = {1 \over 1 - F_2 (\alpha) } f(\epsilon_1 | \epsilon_2 > \alpha) Pr (\epsilon_2 > \alpha )$$

Then: $$ = {1 \over 1 - F_2 (\alpha) } \int_\alpha^\infty f(\epsilon_1, \epsilon_2) d \epsilon_2 $$

$$ = {1 \over 1 - F_2 (\alpha) } \int_\alpha^\infty f(\epsilon_2 | \epsilon_1) f_1 (\epsilon_1) d \epsilon_2 $$

$$ = { f_1 (\epsilon_1) \over 1 - F_2 (\alpha) } \int_\alpha^\infty f(\epsilon_2 | \epsilon_1 ) d \epsilon_2 $$

Focus now on the integral. $f(\epsilon_2 | \epsilon_1 ) $ follows a normal distribution too.

$$ \int_\alpha^\infty f(\epsilon_2 | \epsilon_1 ) d \epsilon_2 = 1 - F_{\epsilon_2 | \epsilon_1} (\alpha) $$

Now that we have the whole expression we just need to convert things into standard normal PDFs and CDFs.

We can write $ f_1 (\epsilon_1) = {1 \over \sigma_{1}} \phi({\epsilon_1 - \mu_1 \over \sigma_1})$ and $F_2 (\alpha) = \Phi ( {\alpha - \mu_2 \over \sigma_2 })$

Now note that $ f(\epsilon_2 | \epsilon_1 )$ is $N(\mu_{2|1}, \sigma_{2|1}) $, where $\mu_{2|1} = \mu_2 + \rho \frac{\sigma_2}{\sigma_1} (\epsilon_1 - \mu_1)$ and $ \sigma_{2|1} = (1 - \rho^2)\sigma_2^2 $, where $\rho$ is the correlation between $\epsilon_1$ and $\epsilon_2$ (check this entry in wikipedia)

Thus, the final expression is as follows:

$$ f(\epsilon_1 | \epsilon_2 > \alpha ) = { \phi\left({\epsilon_1 - \mu_1 \over \sigma_1}\right) \over \sigma_{1} \left( 1 - \Phi \left( {\alpha - \mu_2 \over \sigma_2 }\right) \right)} \left( 1 -\Phi\left( {\alpha - \mu_{2|1} \over \sigma_{2|1}} \right) \right) $$

I might add this result to the bivariate normal entry in wikipedia so that the next poor soul like me that has this doubt can easily find it. For that to happen, please help me check that the steps I followed are correct.

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