Common Distributions - Key Proof

We prove that the sum of independent normal random variables is normal—a fundamental result with far-reaching consequences.

Normality of Sums

Theorem

If $X \sim \mathcal{N}(\mu_1, \sigma_1^2)$ and $Y \sim \mathcal{N}(\mu_2, \sigma_2^2)$ are independent, then: $X + Y \sim \mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$

Proof

We prove using moment generating functions.

Step 1: Recall the MGF of $X \sim \mathcal{N}(\mu, \sigma^2)$ : $M_X(t) = E[e^{tX}] = e^{\mu t + \sigma^2 t^2/2}$

Step 2: For independent $X$ and $Y$ : $M_{X+Y}(t) = E[e^{t(X+Y)}] = E[e^{tX}]E[e^{tY}] = M_X(t) M_Y(t)$

by independence.

Step 3: Substituting the normal MGFs: $M_{X+Y}(t) = e^{\mu_1 t + \sigma_1^2 t^2/2} \cdot e^{\mu_2 t + \sigma_2^2 t^2/2}$

$= e^{(\mu_1 + \mu_2)t + (\sigma_1^2 + \sigma_2^2)t^2/2}$

Step 4: This is the MGF of $\mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$ .

Since MGFs uniquely determine distributions, we conclude: $X + Y \sim \mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$ □

■

Generalization

Theorem

If $X_1, \ldots, X_n$ are independent with $X_i \sim \mathcal{N}(\mu_i, \sigma_i^2)$ , and $a_1, \ldots, a_n$ are constants, then: $\sum_{i=1}^n a_i X_i \sim \mathcal{N}\left(\sum_{i=1}^n a_i \mu_i, \sum_{i=1}^n a_i^2 \sigma_i^2\right)$

Proof Outline: First show $aX \sim \mathcal{N}(a\mu, a^2\sigma^2)$ using MGFs: $M_{aX}(t) = E[e^{taX}] = M_X(at) = e^{\mu(at) + \sigma^2(at)^2/2} = e^{(a\mu)t + (a^2\sigma^2)t^2/2}$

Then apply the sum result iteratively. □

Example

Portfolio Returns: Consider a portfolio with weights $w_1 = 0.6, w_2 = 0.4$ in two stocks.

Stock 1: annual return $R_1 \sim \mathcal{N}(0.08, 0.04^2)$ (8% mean, 4% std dev)

Stock 2: annual return $R_2 \sim \mathcal{N}(0.12, 0.06^2)$ (12% mean, 6% std dev)

Assuming independence, portfolio return: $R_p = 0.6R_1 + 0.4R_2 \sim \mathcal{N}(\mu_p, \sigma_p^2)$

where: $\mu_p = 0.6(0.08) + 0.4(0.12) = 0.096 = 9.6\%$ $\sigma_p^2 = (0.6)^2(0.04)^2 + (0.4)^2(0.06)^2 = 0.000576 + 0.000576 = 0.001152$ $\sigma_p = \sqrt{0.001152} \approx 0.0339 = 3.39\%$

Alternative Proof via Convolution

The PDF of $Z = X + Y$ is the convolution: $f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z-x) \, dx$

For normal densities, this integral evaluates to another normal density (tedious calculation involving completing the square in the exponent).

Remark

The closure of the normal distribution under linear combinations is unique among continuous distributions. This property, combined with the CLT, makes the normal distribution central to probability and statistics. The MGF proof is elegant and generalizes easily to $n$ variables.