Why is the F-Statistic $\approx$ 1 when the null hypothesis is true? I'm currently reading through the section on linear regression in ISLR, and the authors use the F-Statistic to determine if we should reject the null hypothesis (well they use the p-value of the F-Statistic).
I understand that, when the null hypothesis is false, the total sum of squares (the sum of squares where the response is just a constant value) will be much larger than the $RSS$, since our model will explain a large amount of the variation. Therefore, we should see a large F-Statistic value. However, I don't quite understand why, when the null hypothesis is true, we should expect to see a a F-Statistic $\approx 1$. The numerator, from my understanding, should be small, since not much of the variation is explained by our model. However, why should the denominator, which is $RSS/(n-p-1)$ be the same as the numerator? What does the denominator actually represent?
 A: Consider a linear model $y_i=\beta_0+x_i'\beta+u_i$, with $u_i\sim (0,\sigma^2)$.
The F-statistic is (see e.g. Proof that F-statistic follows F-distribution)
$$ F = \frac{(\text{TSS}-\text{RSS})/p}{\text{RSS}/(n-p-1)},$$
with $TSS=\sum_i(y_i-\bar{y})^2$ and $RSS=\sum_i(y_i-\hat{y}_i)^2$ with $p$ the number of slope parameters.
Under classical assumptions, $\text{RSS}/(n-p-1)$ is an unbiased estimator of $\sigma^2$, i.e.,
$$E[\text{RSS}/(n-p-1)]=\sigma^2.$$
Likewise, it is a well-known result that, under the null $y_i=\beta_0+u_i$, the sample variance $\sum_i(y_i-\bar{y})^2/(n-1)$ is an unbiased estimator of $\sigma^2$, i.e.,
$$E[\text{TSS}/(n-1)]=\sigma^2.$$
(It always is an unbiased estimator of the variance of $\sigma^2_y$, the variance of $y$, which however does not coincide with the variance of the error under the alternative anymore, which is what gives the test its power.)
Putting things in the numerator together,
$$
E[(\text{TSS}-\text{RSS})/p]=E[(n-1)\sigma^2-(n-p-1)\sigma^2]/p=\sigma^2
$$
So if you approximate $E(F)$ (of course, the expectation of a ratio is not generally the ratio of expectations), you get
$$
E(F)\approx\frac{E[(\text{TSS}-\text{RSS})/p]}{E[\text{RSS}/(n-p-1)]}=\frac{\sigma^2}{\sigma^2}=1
$$
Actually, given that the F-statistic follows an F-distribution with $p$ and $d:=n-p-1$ degrees of freedom, we may use known results for the exact expectation of F-distributed random variables, namely that
$$E(F)=\frac{d}{d-2}$$
when $d>2$. So
$$E(F)=\frac{n-p-1}{n-p-1-2}=\frac{n-p-1}{n-p-3},$$
which will of course be close to 1 for cases where the sample size $n$ is large relative to the number of regressors. Hence, the above approximation works very well in this case.
Of course, what we have here is a result for the expected value of the F-statistic if the null is true. This does not mean that (like for any expectation) the statistic $F\approx1$, but that it will "hover around" 1 when we were to repeatedly compute F-statistic for situations in which the null is true. See e.g. the simulation provided at https://stats.stackexchange.com/a/258476/67799 for an illustration.
A: Another way to think about this:
When there is no experimental effect, the only variation you have is within-subjects.  So even when you split the participants into two groups, the same variation is endemic within each (no experimental variance, only within-subjects exists).  Thus, your variance ratio of between / within is really just within / within, or 1.
