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If I have two normally distributed random variables $a\sim\mathcal{N}\left(\mu_a,\sigma_a\right)$ and $b\sim\mathcal{N}\left(\mu_b,\sigma_b\right)$ with correlation $\rho$, is there a closed form for the correlation between $a$ and a transformation of $b$: $$ b^\prime=\begin{cases} b & \text{if } b\leq\mu_b \\ (1-\eta) b+\eta\mu_b & \text{ if } b>\mu_b \end{cases}, $$ for some $\eta\in\left[0,1\right]$.

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If we assume joint normality (here, it doesn't seem possible to calculate the correlation coef. when we don't have information about the joint distribution), we can use bivariate normal distribution for formulating the joint PDF. We need $cov(a,b')$ and $var(b')$.

First, we write $b'$ in terms of indicators as $b'=b\mathcal{I}(b\leq\mu_b)+((1-\eta)b+\eta\mu_b)\mathcal{I}(b>\mu_b)$. Since $\mathcal{I}(b>\mu_b)=1-\mathcal{I}(b\leq\mu_b)$, $b'$ becomes $b'=\eta(b-\mu_b)\mathcal{I}(b\leq\mu_b)+(1-\eta)(b-\mu_b)+\mu_b$. The final term $\mu_b$ is constant and doesn't affect both variance and covariance calculations.

$$\begin{align}cov(a,b') &=\eta cov(a,(b-\mu_b)I(b\leq\mu_b))+(1-\eta)cov(a,b)\\ &= \eta cov(v,u\mathcal{I}(u\leq 0))+(1-\eta)\rho\sigma_a\sigma_b \\ &= \eta E[vu\mathcal{I}(u\leq 0)]+(1-\eta)\rho\sigma_a\sigma_b\end{align}$$ where $u=b-\mu_b$, $v=a-\mu_a$ (de-meaning $a$ doesn't change the covariance). Using joint PDF, ($C=\frac{1}{2\pi\sigma_a\sigma_b\sqrt{1-\rho^2}}$), we can calculate the above expectation term as follows:

$$\begin{align}E[vu\mathcal{I}(u\leq 0)] & = C\int_{-\infty}^0 u\int_{-\infty}^\infty v\exp\left({-\frac{1}{2(1-\rho^2)}\left(\frac{u^2}{\sigma_b^2}-\frac{2\rho uv}{\sigma_a\sigma_b}+\frac{v^2}{\sigma_a^2}\right)}\right)dvdu \\ &= C\int_{-\infty}^0u\exp\left(-\frac{u^2}{2\sigma_b^2}\right)\int_{-\infty}^{\infty}v\exp\left(-\frac{1}{2(1-\rho^2)\sigma_a^2}\left(v-\frac{u\rho\sigma_a}{\sigma_b}\right)^2\right)dvdu \\ &= \frac{1}{2\pi\sigma_a\sigma_b\sqrt{1-\rho^2}}\int_{-\infty}^0u\exp\left(-\frac{u^2}{2\sigma_b^2}\right)\sqrt{2\pi}\sigma_a\sqrt{1-\rho^2}\frac{u\rho\sigma_a}{\sigma_b}du \\ &= \frac{\rho\sigma_a}{\sqrt{2\pi}\sigma_b^2}\frac{1}{2}\int_{-\infty}^{\infty}u^2\exp\left(-\frac{u^2}{2\sigma_b^2}\right)du=\frac{\rho\sigma_a\sigma_b}{2}\end{align}$$

When we substitute: $$cov(a,b')=\boxed{\frac{(2-\eta)}{2}\rho\sigma_a\sigma_b}$$

For the variance, we use a similar method with variable changes and ignoring constant terms: $$\begin{align}var(b') &=cov(b',b')=cov(\eta u \mathcal{I}(u\leq 0)+(1-\eta)u,\eta u \mathcal{I}(u\leq 0)+(1-\eta)u) \\ &= (2\eta-\eta^2)E[u^2\mathcal{I}(u\leq 0)]-\eta^2E[u\mathcal{I}(u\leq 0)]^2+(1-\eta)^2E[u^2]\end{align}$$

Calculating the terms: $$E[u^2\mathcal{I}(u\leq 0)]=\int_{-\infty}^0u^2f(u)du=\frac{1}{2}E[u^2]=\frac{\sigma_b^2}{2}$$ $$E[u\mathcal{I}(u\leq 0)]=\int_{-\infty}^0u\frac{1}{\sqrt{2\pi}\sigma_b}\exp\left(\frac{-u^2}{2\sigma_b^2}\right)du=-\frac{\sigma_b}{\sqrt{2\pi}} \text{when } \alpha=\frac{u^2}{2\sigma_b^2}$$ Substituting yields the following weird formula: $$var(b')=\sigma_b^2\left(\frac{(1-\eta)^2+1}{2}-\frac{\eta^2}{2\pi}\right)\rightarrow \rho_{a,b'}=\boxed{\rho\frac{2-\eta}{\sqrt{4-4\eta+2\eta^2(1-1/\pi)}}}$$

which changes in $\approx [0.856\rho,\rho]$.

Note: numerically verified.

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