• Always positive and right-skewed
• $F_{d_1, d_2}$ is related to the reciprocal: $1/F_{d_1, d_2} = F_{d_2, d_1}$
• As degrees of freedom increase, F approaches 1
• The mode (for $d_2 > 2$) is at $\frac{d_1 - 2}{d_1} \cdot \frac{d_2}{d_2 + 2} < 1$

The variance exists only when $d_2 > 4$. The F-distribution is always right-skewed, with skewness decreasing as both degrees of freedom increase.

When $d_1 = d_2$, $E[F] = d/(d-2)$, which approaches 1 as $d \to \infty$. This makes sense: if both variances estimate the same $\sigma^2$, their ratio should be near 1.

Step 1. By the chi-square result: $U = \frac{(n_1-1)s_X^2}{\sigma^2} \sim \chi^2_{n_1-1}$ and $V = \frac{(n_2-1)s_Y^2}{\sigma^2} \sim \chi^2_{n_2-1}$.

Step 2. Since the samples are independent, $U$ and $V$ are independent.

Step 3. Therefore $F = \frac{U/(n_1-1)}{V/(n_2-1)} = \frac{s_X^2}{s_Y^2} \sim F_{n_1-1, n_2-1}$ by definition.

The key requirements are: (1) normal populations, (2) equal variances, and (3) independent samples.

Under $H_0$, all observations come from $N(\mu, \sigma^2)$. The numerator $\text{MS}_{\text{between}}$ is a chi-square variable divided by $k-1$, and $\text{MS}_{\text{within}}$ is a chi-square variable divided by $N-k$, and they are independent by Cochran's theorem. Their ratio follows $F_{k-1, N-k}$.