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Here is the problem I'm trying to work out: Let $v_b,v_s$ be jointly normally distributed random variables with pdf $f(v_b,v_s)$. I want to work out $E[v_b|v_s\leq\pi]$ for some constant $\pi$. Here is what I tried:

$$E[v_b|v_s\leq\pi]=\frac{1}{F_{v_s}(\pi)}\int_{-\infty}^{\infty}v_bf_{v_b|v_s\leq\pi}dv_b=\frac{1}{F_{v_s}(\pi)}\int_{-\infty}^{\infty}v_b\frac{f_{v_b}}{F_{v_s}(\pi)}dv_b=\frac{1}{F_{v_s}(\pi)^2}E[v_b]$$

However, I'm not sure I should be integrating from $-\infty$ to $\infty$. Is this correct?

Dmlawton
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  • See [this site search](https://stats.stackexchange.com/search?q=normal+conditional+trunc*+score%3A2). – whuber Mar 02 '21 at 18:47
  • Since the event $v_s\le \pi$ is the event $\pi-v_s \ge 0,$ which has the same probability as $\pi-v_s\gt 0,$ and because $(v_b, \pi-v_s)$ has a bivariate Normal distribution whose parameters are obvious, the duplicate question directly applies. – whuber Mar 02 '21 at 18:50

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