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Let $U,V$ be zero-mean normally distributed with covariances $\sigma_u^2,\sigma_v^2,\sigma_{u,v}$.

Problem: Show that: $$ E[U|U-V>c]=\frac{\sigma_{u,u-v}}{\sigma_{u-v}}\cdot\frac{f\left(\frac{c}{\sigma_{u-v}}\right)}{1-F\left(\frac{c}{\sigma_{u-v}}\right)} $$ where $\sigma_{u,u-v}$ is the covariance between $U$ and $U-V$, and $f,F$ are the pdf and CDF of the standard normal.

What I Know: I have this result for $Z\sim N(0,1)$: $$ E[Z|Z>c]=\frac{f(c)}{1-F(c)} $$

Main Question: Do you have any tips on how I could use the property above to solve the Problem? Thanks for helping! :D

Guilherme Salomé
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    Hint: the form of the answer strongly suggests changing to a new variable $W=U-V$ and standardizing it. At that point, drawing a picture in $(W,U)$ coordinates should make it apparent how to reduce this question to one about a univariate standard Normal variable. – whuber Mar 31 '17 at 16:07
  • I solved it! Thanks @whuber. I'll type the answer and post it here. – Guilherme Salomé Mar 31 '17 at 16:46

1 Answers1

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Let's begig using @whuber's suggestion: $$ \begin{align*} E[U|\underbrace{U-V}_{\equiv W}>c]&=E\left[\frac{U}{\sigma_U}\cdot\sigma_U\bigg\vert\frac{W}{\sigma_W}>\frac{c}{\sigma_W}\right]\\ &=\sigma_UE\left[\frac{U}{\sigma_U}\bigg\vert\frac{W}{\sigma_W}>\frac{c}{\sigma_W}\right]\\ &=\sigma_U\int_{c/\sigma_W}^{+\infty}E\left[\frac{U}{\sigma_U}\bigg\vert\frac{W}{\sigma_W}=a\right]f_{W/\sigma_W}(a)da \end{align*} $$ Now, because both terms inside the expectation above are standard normal, we have: $$ \begin{align*} E\left[\frac{U}{\sigma_U}\bigg\vert\frac{W}{\sigma_W}=a\right]&=E\left[\frac{U}{\sigma_U}\right]+Cov\left(\frac{U}{\sigma_U},\frac{W}{\sigma_W}\right)\cdot Var\left(\frac{W}{\sigma_W}\right)\cdot\left(a-E\left[\frac{W}{\sigma_W}\right]\right)\\ &=\frac{\sigma_{U,W}}{\sigma_U\sigma_W}\cdot a \end{align*} $$ (This property was what I was missing. I couldn't figure this out visually though =/, any help here would be appreciated :D) $$ \begin{align*} E[U|U-V>c]&=\sigma_U\int_{c/\sigma_W}^{+\infty}E\left[\frac{U}{\sigma_U}\bigg\vert\frac{W}{\sigma_W}=a\right]f_{W/\sigma_W}(a)da\\ &=\sigma_U\frac{\sigma_{U,W}}{\sigma_U\sigma_W}\int_{c/\sigma_W}^{+\infty}af_{W/\sigma_W}(a)da\\ &=\frac{\sigma_{U,W}}{\sigma_W}E\left[\frac{W}{\sigma_W}\bigg\vert\frac{W}{\sigma_W}>\frac{c}{\sigma_W}\right]\\ &=\frac{\sigma_{U,W}}{\sigma_W}\cdot\frac{f\left(\frac{c}{\sigma_W}\right)}{1-F\left(\frac{c}{\sigma_W}\right)} \end{align*} $$ where the last line used the result for the standard normal mentioned in the problem.

Since $W=U-V$, $\sigma_{U,W}=\sigma_{U,U-V},\sigma_W=\sigma_{U-V}$, and we have the solution!

Guilherme Salomé
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