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I have two jointly normally distributed variables $s_1$ and $s_2$. I am now searching for the conditional expectation $$ E(s_1|s_1>r_1,\ s_2>r_2) $$ where $r_1$ and $r_2$ are constants. An idea of how to arrive at this and sources where it can be read up would also be greatly appreciated.

gung - Reinstate Monica
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Mike
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  • What do you know about the *joint* distribution of $s_1$ and $s_2$? This information is essential for determining the conditional expectation. – whuber Jun 12 '14 at 14:21
  • $s_1$ and $s_2$ are generally correlated, but assume they have unconditional mean zero and variance one. If they weren't correlated, it'd be easy I think... – Mike Jun 12 '14 at 14:35
  • Is the joint distribution bivariate Normal? – whuber Jun 12 '14 at 14:37
  • Yes, let's say: $$s_1,s_2 \sim \mathcal{N}(0,1) $$ and the correlation is $\rho$. Sorry, I don't manage to type them together in one bivariate normal... – Mike Jun 12 '14 at 14:48
  • Unfortunately, both integrals in the ratio that answers this question are intractable when $\rho\ne 0$ and $\rho \ne \pm 1$ and have to be computed numerically. – whuber Jun 12 '14 at 16:42
  • Thanks. I meanwhile found that $$E(s_1|s_1>r_1)=\frac{\phi(r_1)}{1-\Phi(r_1)}$$ and $$E(s_1|s_2>r_2)=\rho\frac{\phi(r_2)}{1-\Phi(r_2)}.$$ But you say that we cannot write $ E(s_1|s_1>r_1,\ s_2>r_2)$ along these lines? – Mike Jun 12 '14 at 18:47
  • I would also be surprised if we can not write $E(s_1|s_1>r_1,\ s_2>r_2)$ in the same way. – AnonA Feb 18 '21 at 04:22
  • Any intuition on why this conditional expectation is different would be appreciated – AnonA Feb 18 '21 at 08:30

1 Answers1

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To summarize the comments: Since we assume that $s_1$ and $s_2$ jointly follow a standard bivariate normal distribution, with correlation coefficient $\rho$, then the joint density is

$$f(s_1,s_2) = \frac{1}{2 \pi \sqrt{1-\rho^2}} \exp\left\{-\frac{s_1^2 +s_2^2 -2\rho s_1s_2}{2(1-\rho^2)}\right\} $$

We also have

$$E(s_1|s_1>r_1,\ s_2>r_2) = \frac {E(s_1;\{s_1>r_1,\ s_2>r_2\})}{P(s_1>r_1,\ s_2>r_2)}$$

$$=\frac {\int_{r_2}^{\infty}\int_{r_1}^{\infty}s_1f(s_1,s_2)ds_1ds_2}{\int_{r_2}^{\infty}\int_{r_1}^{\infty}f(s_1,s_2)ds_1ds_2} $$

The fact that the conditioning statement includes, and places bounds on, both variables, does not permit us to simplify this ratio of integrals (as would be in the cases described by the OP in the comments). Moreover, as @whuber writes, these integrals do not have an analytical solution for $\rho \ne \{0,\pm1\}$, and must be computed numerically for each $\{r_1, r_2\}$.

Alecos Papadopoulos
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