Joint Normal Conditional Probability
Let $(R_1, R_2)$ be jointly normal with means $\mu_1 = \mu_2 = 0$, standard deviations $\sigma_1 = \sigma_2 = 1\%$, and correlation $\rho = 0.85$.
Compute $P(R_1 > 1\%, \, R_2 < 0)$ by using the conditional distribution of $R_1 \mid R_2$. A numeric approximation is fine -- you can quote standard normal table values.
Hints
- Standardize both returns so you are working with standard normals. The event becomes $P(Z_1 > 1, Z_2 < 0)$ -- what does the conditional distribution $Z_1 \mid Z_2$ look like?
- The conditional distribution is $Z_1 \mid Z_2 = z \sim N(\rho z, 1 - \rho^2)$. Write out $P(Z_1 > 1 \mid Z_2 = z)$ as a $\Phi$ expression and integrate over $z < 0$.
- The integral $\int_{-\infty}^{0} \Phi\!\left(\frac{\rho z - 1}{\sqrt{1-\rho^2}}\right) \phi(z)\,dz$ can be recognized as the bivariate normal CDF $\Phi_2(-1, 0; -\rho)$. Evaluate numerically by checking the integrand at $z = 0, -0.5, -1$.
Worked Solution
How to Think About It: You have two highly correlated returns. One return is strongly positive (one standard deviation up) while the other is negative. With $\rho = 0.85$, these two returns move together most of the time, so asking for one to be up while the other is down is asking for a rare event. Your gut should say this probability is small -- much smaller than the unconditional $P(R_1 > 1\%) = \Phi(-1) \approx 15.9\%$. Conditioning on $R_2 < 0$ drags the conditional mean of $R_1$ negative, making $R_1 > 1\%$ very unlikely.
Quick Estimate: Standardize by setting $Z_i = R_i / \sigma$. Then $(Z_1, Z_2)$ is standard bivariate normal with $\rho = 0.85$, and the event becomes $P(Z_1 > 1, Z_2 < 0)$. If $Z_2 < 0$, a rough representative value is $E[Z_2 \mid Z_2 < 0] = -\phi(0)/\Phi(0) \approx -0.80$. At that point the conditional mean of $Z_1$ is $0.85 \times (-0.80) = -0.68$, and the conditional standard deviation is $\sqrt{1 - 0.85^2} \approx 0.527$. The z-score for $Z_1 > 1$ becomes $(1 - (-0.68))/0.527 \approx 3.19$, giving $P \approx \Phi(-3.19) \approx 0.07\%$. This is only a rough proxy (we used the truncated mean instead of integrating), but it tells us the answer is on the order of a few tenths of a percent.
Approach: Use the law of total probability by conditioning on $R_2$, then integrate over the region $R_2 < 0$.
Formal Solution:
Standardize: let $Z_1 = R_1/\sigma$ and $Z_2 = R_2/\sigma$, so $(Z_1, Z_2)$ is standard bivariate normal with correlation $\rho = 0.85$. The target is
$$P(Z_1 > 1, \, Z_2 < 0).$$
The conditional distribution of $Z_1 \mid Z_2 = z$ is
$$Z_1 \mid Z_2 = z \;\sim\; N\!\left(\rho \, z, \; 1 - \rho^2\right).$$
So with $\sigma_c = \sqrt{1 - \rho^2} = \sqrt{1 - 0.7225} = \sqrt{0.2775} \approx 0.5268$,
$$P(Z_1 > 1 \mid Z_2 = z) = 1 - \Phi\!\left(\frac{1 - \rho z}{\sigma_c}\right) = \Phi\!\left(\frac{\rho z - 1}{\sigma_c}\right).$$
By the law of total probability,
$$P(Z_1 > 1, \, Z_2 < 0) = \int_{-\infty}^{0} \Phi\!\left(\frac{0.85\,z - 1}{0.5268}\right) \phi(z) \, dz.$$
This integral equals the bivariate normal rectangle probability $\Phi_2(-1, 0; -0.85)$ -- i.e., the standard bivariate normal CDF evaluated at $(-1, 0)$ with correlation $-\rho = -0.85$. To see this, note
$$P(Z_1 > 1, Z_2 < 0) = P(-Z_1 < -1, Z_2 < 0)$$
and $(-Z_1, Z_2)$ is bivariate normal with correlation $-\rho$.
To get a numeric value, evaluate the integrand at a few points:
- $z = 0$: $\Phi((0 - 1)/0.5268) = \Phi(-1.898) \approx 0.0288$
- $z = -0.5$: $\Phi((-0.425 - 1)/0.5268) = \Phi(-2.705) \approx 0.0034$
- $z = -1$: $\Phi((-0.85 - 1)/0.5268) = \Phi(-3.512) \approx 0.00022$
The integrand drops off rapidly, so most of the contribution comes from $z$ near $0$. Numerical integration gives
$$P(Z_1 > 1, Z_2 < 0) \approx 0.0026.$$
Answer: $P(R_1 > 1\%, \, R_2 < 0) \approx 0.26\%$, obtained by integrating $\Phi\!\left(\frac{\rho z - 1}{\sqrt{1-\rho^2}}\right)\phi(z)$ over $z < 0$, which equals $\Phi_2(-1, 0; -\rho) \approx 0.0026$.
Intuition
High correlation acts as a strong leash between two variables. When $\rho = 0.85$, conditioning on $R_2 < 0$ yanks the conditional mean of $R_1$ well below zero, so asking $R_1$ to simultaneously exceed one standard deviation in the opposite direction is fighting the correlation. The conditional standard deviation shrinks by a factor of $\sqrt{1 - \rho^2} \approx 0.53$, so $R_1$ has much less room to wander away from its conditional mean. The combination of a shifted mean and reduced variance crushes the probability from about 16% (unconditional) down to about 0.26%.
This pattern shows up constantly in portfolio risk. If you hold two highly correlated assets, the scenario where one crashes while the other rallies is extremely unlikely -- and that is precisely the hedge you are counting on. Conversely, if you are pricing a digital option on the spread $R_1 - R_2$, the probability of large dislocations depends sensitively on the conditional variance $1 - \rho^2$. This is why traders obsess over correlation: it controls how much diversification you actually get, and tiny changes in $\rho$ near 1 have outsized effects on tail probabilities.