Most Likely Split of Two Normals Given Their Sum
You have two independent normally distributed random variables $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$. You are told that $X + Y = 100$.
What is the most likely value of $X$ (and hence $Y = 100 - X$)?
Work through the special cases:
- $X$ and $Y$ are both standard normal, i.e., $N(0, 1)$.
- $X$ and $Y$ have the same mean but different variances.
- The fully general case with arbitrary means and variances.
Hints
- Conditioning a joint Gaussian on a linear constraint gives another Gaussian. What does the mode of a normal distribution equal?
- Think about how $\text{Cov}(X, X+Y)$ relates to how $X$ adjusts when you learn $X + Y = 100$. The conditional mean formula for bivariate normals will give you the answer directly.
- The conditional mean is $\mu_X + \frac{\sigma_X^2}{\sigma_X^2 + \sigma_Y^2}(100 - \mu_X - \mu_Y)$. Each variable absorbs the gap between 100 and the sum of means in proportion to its variance.
Worked Solution
How to Think About It: Once you know $X + Y = 100$, you are constraining a two-dimensional Gaussian to a line. The "most likely" point on that line is where the original joint density is highest subject to the constraint. For normals, the conditional distribution is also normal, so the mode equals the conditional mean. The intuition is simple: each variable gets pulled toward its own mean, but the one with more variance (more "flexibility") absorbs more of whatever slack is needed to hit the sum of 100. If you are at a whiteboard, start with equal variances -- the answer is obviously symmetric.
Quick Estimate: Suppose $\mu_X = 30$, $\mu_Y = 40$, $\sigma_X^2 = 4$, $\sigma_Y^2 = 1$. The unconstrained means sum to 70, so there is a gap of 30 to distribute. Each variable should absorb the gap in proportion to its variance. $X$ has $4/5$ of the total variance, $Y$ has
Approach: Derive the conditional distribution $X \mid (X + Y = 100)$ using the joint Gaussian structure, then read off the mode (which equals the mean for a normal).
Formal Solution:
Since $X$ and $Y$ are independent normals, $X + Y \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$. The joint distribution of $(X, X+Y)$ is bivariate normal. The conditional distribution of $X$ given $X + Y = s$ is:
$X \mid (X + Y = s) \sim N\!\left(\mu_X + \frac{\sigma_X^2}{\sigma_X^2 + \sigma_Y^2}(s - \mu_X - \mu_Y),\; \frac{\sigma_X^2 \sigma_Y^2}{\sigma_X^2 + \sigma_Y^2}\right)$
With $s = 100$, the conditional mean (and mode) of $X$ is:
$X^{*} = \mu_X + \frac{\sigma_X^2}{\sigma_X^2 + \sigma_Y^2}(100 - \mu_X - \mu_Y)$
and $Y^{*} = 100 - X^{*}$, which by symmetry of the derivation gives:
$Y^{*} = \mu_Y + \frac{\sigma_Y^2}{\sigma_X^2 + \sigma_Y^2}(100 - \mu_X - \mu_Y)$
Special cases:
- Standard normal ($\mu_X = \mu_Y = 0$, $\sigma_X^2 = \sigma_Y^2 = 1$): Equal variances, so the gap of 100 is split evenly. $X^{*} = Y^{*} = 50$.
- Same mean, different variances ($\mu_X = \mu_Y = \mu$): The formula simplifies to $X^{*} = \mu + \frac{\sigma_X^2}{\sigma_X^2 + \sigma_Y^2}(100 - 2\mu)$. The variable with the larger variance absorbs a larger share of the deviation from
- General case: Use the full formula above. Each variable starts at its own mean and then adjusts in proportion to its share of total variance.
Answer: The most likely value of $X$ given $X + Y = 100$ is:
$X^{*} = \mu_X + \frac{\sigma_X^2}{\sigma_X^2 + \sigma_Y^2}(100 - \mu_X - \mu_Y)$
with $Y^{*} = 100 - X^{*}$. For identical distributions, $X^{*} = Y^{*} = 50$. The key principle: each variable absorbs the slack in proportion to its variance.
Intuition
This problem illustrates a fundamental property of Gaussians: conditioning on a linear constraint preserves normality, and the adjustment from the prior mean is proportional to the variable's share of total variance. In a portfolio context, this is exactly how you decompose a known total P&L across desks or positions -- the desk with more volatile returns "explains" more of the surprise. It is also the logic behind BLUE (best linear unbiased estimation) and Kalman filtering: new information gets distributed to each component in proportion to its uncertainty.
The common mistake is to say "they split 50/50" without checking whether the variances are equal. Equal split only holds for identical distributions. When variances differ, the more uncertain variable does the heavy lifting -- it moves further from its mean because the data is less surprising for it. This variance-weighted allocation shows up everywhere in quant work: risk budgeting, signal combination, and Bayesian shrinkage estimators.