Quadrant Probabilities for Standard Bivariate Normals

Why Zero-Threshold Quadrants Have a Simple Formula

In a bivariate normal model, the dependence between two measurements affects how often they have the same sign. Positive correlation makes simultaneous positive values more likely; negative correlation makes opposite signs more likely. At zero thresholds, this relationship has an especially concise form: a quadrant probability is a constant plus or minus an arcsine of the correlation.

The result is useful when a problem asks whether two centered measurements are both above their means, whether they have opposite signs, or how sign agreement changes with correlation. The thresholds must genuinely be zero after standardization. A correlation value alone is not enough to use this formula for arbitrary thresholds.

Definition: Let $X$ and $Y$ have a standard bivariate normal distribution with $E[X]=E[Y]=0$, $\operatorname{Var}(X)=\operatorname{Var}(Y)=1$, and $\operatorname{Corr}(X,Y)=\rho$, where $-1<\rho<1$. Then $P(X>0,Y>0)$ is the upper-right quadrant probability. The other three quadrants are defined by the signs of $X$ and $Y$. Because this model is continuous, replacing a strict inequality by a non-strict one does not change these probabilities.

Deriving the Arcsine Formula

A useful representation of a standard bivariate normal pair is $X=U$ and $Y=\rho U+\sqrt{1-\rho^2}\,V$, where $U$ and $V$ are independent standard normal random variables. This representation has standard normal marginals and correlation $\rho$. Write $s=\sqrt{1-\rho^2}$; under the stated assumption $-1<\rho<1$, $s>0$.

The pair $(U,V)$ has density $(2\pi)^{-1}\exp(-(u^2+v^2)/2)$. In polar coordinates $U=R\cos\theta$, $V=R\sin\theta$, this density depends on $R$ but not on $\theta$. Thus the angle is uniform over a full circle: an angular region of length $L$ has probability $L/(2\pi)$. The event $X=0$ or $Y=0$ has probability zero, so the origin and the quadrant boundaries do not affect the calculation.

The condition $X>0$ becomes $\cos\theta>0$, or $-\pi/2<\theta<\pi/2$. Define $\alpha=\arccos(\rho)$, so that $\cos\alpha=\rho$ and $\sin\alpha=\sqrt{1-\rho^2}$. Then $Y=R\{\rho\cos\theta+\sqrt{1-\rho^2}\sin\theta\} =R\cos(\theta-\alpha)$. Consequently, $Y>0$ when $\alpha-\pi/2<\theta<\alpha+\pi/2$, with angles interpreted around the circle. The overlap with the interval for $X>0$ has length $\pi-\alpha$.

P(X>0,Y>0) =\frac{\pi-\alpha}{2\pi} =\frac{\pi-\arccos(\rho)}{2\pi} =\frac14+\frac{\arcsin(\rho)}{2\pi}.

The last equality uses $\arcsin(\rho)+\arccos(\rho)=\pi/2$ for $-1<\rho<1$, with $\arcsin$ taking values in $[-\pi/2,\pi/2]$. The angular overlap argument gives a geometric reason for the formula: changing $\rho$ changes the angle between the two half-plane conditions, and probability is proportional to the resulting angular width.

Theorem: For a standard bivariate normal pair with $-1<\rho<1$, $$ P(X>0,Y>0)=P(X<0,Y<0)=\frac14+\frac{\arcsin(\rho)}{2\pi}, $$ and $$ P(X>0,Y<0)=P(X<0,Y>0)=\frac14-\frac{\arcsin(\rho)}{2\pi}. $$

The second and third expressions follow from symmetry and the marginal probabilities, not from a separate integration. Central symmetry of the pair gives equal probabilities for the upper-right and lower-left quadrants. Also, $P(X>0)=1/2$, so subtracting $P(X>0,Y>0)$ from $1/2$ gives $P(X>0,Y<0)$. Central symmetry gives the same probability for the other opposite-sign quadrant.

P(\text{same signs})=\frac12+\frac{\arcsin(\rho)}{\pi}, \qquad P(\text{opposite signs})=\frac12-\frac{\arcsin(\rho)}{\pi}.

These expressions add to one, as they must. Conditional probabilities for a sign event follow from the definition of conditional probability. For example, since $P(X>0)=1/2$, $P(Y>0\mid X>0)=2P(X>0,Y>0) =1/2+\arcsin(\rho)/\pi$. This is a statement about conditioning on the event $X>0$, not conditioning on a particular value of $X$.

Worked Applications

Worked Example: A Positively Associated Pair

Suppose $X$ and $Y$ are standard bivariate normal with correlation $\rho=1/2$. Find the probability that they have the same sign, the probability that they have opposite signs, and $P(Y>0\mid X>0)$.

Method and assumptions. The variables are centered, have unit variances, are jointly normal, and the thresholds are zero, so the arcsine quadrant formula applies. Since $\arcsin(1/2)=\pi/6$,

P(X>0,Y>0)=\frac14+\frac{\pi/6}{2\pi} =\frac14+\frac1{12} =\frac13.

By central symmetry, $P(X<0,Y<0)=1/3$. Each opposite-sign quadrant has probability $\frac14-\frac1{12}=\frac16$. Therefore, the probability of the same sign is $1/3+1/3=2/3$, and the probability of opposite signs is $1/6+1/6=1/3$.

For the conditional probability, divide by $P(X>0)=1/2$: $P(Y>0\mid X>0)=(1/3)/(1/2)=2/3$. The four quadrants sum to $1/3+1/3+1/6+1/6=1$, which checks the calculation. The common mistake avoided here is treating a quadrant probability as the probability of the entire same-sign event; the latter includes two quadrants.

Worked Example: Strong Negative Correlation

A hypothetical monitoring system records two centered, standardized errors $X$ and $Y$. Assume they are jointly normal with correlation $\rho=-\sqrt{3}/2$. Find the probability that both errors are positive, the probability that exactly one is positive, and the conditional probability $P(Y>0\mid X>0)$.

Method and assumptions. The model is a standard bivariate normal pair with zero thresholds and $-1<\rho<1$. Since $\arcsin(-\sqrt{3}/2)=-\pi/3$,

P(X>0,Y>0)=\frac14+\frac{-\pi/3}{2\pi} =\frac14-\frac16 =\frac1{12}.

Exactly one error is positive in either opposite-sign quadrant. Each has probability $\frac14-\frac{-\pi/3}{2\pi}=\frac14+\frac16=\frac5{12}$. Thus the probability that exactly one is positive is $5/12+5/12=5/6$. The conditional probability is $(1/12)/(1/2)=1/6$.

As a check, the two same-sign quadrants each have probability $1/12$, so their total is $1/6$; adding $5/6$ for opposite signs gives one. The negative correlation makes opposite signs much more likely than matching signs, which is consistent with the result.

Worked Example: A Correlation Close to One

For a hypothetical pair of standardized, jointly normal measurements, let $\rho=0.99$. Find the probability that both measurements are positive and the probability that exactly one is positive. Report probabilities to four decimal places.

Method and assumptions. Both variables have mean zero and variance one, the pair is jointly normal, and the thresholds are zero. A calculator gives $\arcsin(0.99)\approx1.4292569$ radians. Using radians,

P(X>0,Y>0) =\frac14+\frac{1.4292569}{2\pi} \approx \frac14+0.2274733 =0.4774733 \approx 0.4775.

Each opposite-sign quadrant has probability $\frac14-0.2274733=0.0225267$. The probability that exactly one measurement is positive is therefore $2(0.0225267)=0.0450534\approx0.0451$.

The four quadrant probabilities are $0.4774733,\ 0.4774733,\ 0.0225267,$ and $0.0225267$; their sum is $1.0000000$, checking the result before rounding. The common mistake avoided is interpreting a correlation of $0.99$ as making the two signs identical with certainty. The probability of opposite signs is small, but it is not zero.

Checks, Limits, and Scope

At $\rho=0$, the formula gives $P(X>0,Y>0)=1/4$. For a bivariate normal pair, zero correlation implies independence, so this agrees with $P(X>0)P(Y>0)=(1/2)(1/2)=1/4$. This is a useful check, but the independence conclusion depends on joint normality; zero correlation alone does not imply independence for an arbitrary pair of random variables.

As $\rho$ approaches $1$ from below, $\arcsin(\rho)$ approaches $\pi/2$, and the upper-right quadrant probability approaches $1/2$. As $\rho$ approaches $-1$ from above, it approaches zero. These are limiting checks for the nondegenerate formula. At the exact endpoints, the bivariate normal pair is degenerate and must be treated under its endpoint model rather than by relying on the nondegenerate density argument.

The formula also has a specific threshold requirement. If $X$ and $Y$ are jointly normal with means $\mu_X,\mu_Y$ and positive standard deviations $\sigma_X,\sigma_Y$, then the formula applies to the events $X>\mu_X$ and $Y>\mu_Y$, after standardizing. It does not directly give $P(X>0,Y>0)$ when the means are nonzero, because zero may then be a nonzero standardized threshold. That case requires a joint probability calculation at the corresponding standardized thresholds.

Key takeaway: For a standard bivariate normal pair at zero thresholds, the upper-right probability is $1/4+\arcsin(\rho)/(2\pi)$. The lower-left quadrant has the same probability, while each opposite-sign quadrant has $1/4-\arcsin(\rho)/(2\pi)$. Check the model, the thresholds, the angle units, and whether the question asks for one quadrant or a pair of quadrants.

Common Exam Traps

Using the formula at nonzero standardized thresholds. The arcsine expression is for the sign quadrants of a centered standard pair. A nonzero threshold changes the region and needs a different calculation.
Dropping the factor of two. $P(X>0,Y>0)$ is one quadrant. The probability of matching signs includes both the upper-right and lower-left quadrants.
Changing the sign in the opposite quadrant. For $\rho>0$, a same-sign quadrant has probability above $1/4$; an opposite-sign quadrant has probability below $1/4$.
Using degrees instead of radians. The formula uses $\arcsin(\rho)$ in radians because it is divided by $2\pi$. Calculator angle mode must be checked.
Assuming zero correlation always means independence. That implication holds for a jointly normal pair, not for every distribution.

Check Your Understanding

Use the zero-threshold formulas, and keep track of whether each question asks for a single quadrant or a union of quadrants.

A standard bivariate normal pair has $\rho=0.6$. Find the probability both variables are positive and the probability exactly one is positive. (Answer: $P(X>0,Y>0)\approx0.3524$; $P(\text{exactly one positive})\approx0.2952$.)
A standard bivariate normal pair has $\rho=-0.5$. Find the probability both variables are negative and the probability of opposite signs. (Answer: $P(X<0,Y<0)=1/6$; $P(\text{opposite signs})=2/3$.)
A standard bivariate normal pair has $\rho=\sqrt{2}/2$. Find $P(Y>0\mid X>0)$. (Answer: $3/4$.)
A standard bivariate normal pair has $\rho=-0.8$. Find the probability both variables are positive and the probability exactly one is positive. Report to four decimal places. (Answer: $P(X>0,Y>0)\approx0.1024$; $P(\text{exactly one positive})\approx0.7952$.)
A standard bivariate normal pair has $\rho=0.999$. Use a calculator in radians to find the probability both variables are positive and the probability of opposite signs. Report to four decimal places. (Answer: $P(X>0,Y>0)\approx0.4929$; $P(\text{opposite signs})\approx0.0142$.)

What You'll Learn