12.2.1. The Mean as a Least Squares Predictor#

What we have shown is the predictor ${\mu }_X$ has the smallest mean squared error among all choices $c$. That smallest mean squared error is the variance of $X$, and hence the smallest root mean squared error is the SD ${\sigma }_X$.

This is why a common approach to prediction is, “My guess is the mean, and I’ll be off by about an SD.”

12.2.2. German Tanks, Revisited#

Recall the German tanks problem in which we have a sample $X_1,X_2,\dots ,X_n$ drawn at random without replacement from $1,2,\dots ,N$ for some fixed $N$, and we are trying to estimate $N$.

We came up with two unbiased estimators of $N$:

• An estimator based on the sample mean: $T_1=2{\overline{X}}_n-1$ where ${\overline{X}}_n$ is the sample average $\frac{1}{n}\sum _{i=1}^n{X}_i$

• An estimator based on the sample maximum: $T_2=M\cdot \frac{n+1}{n}-1$ where $M=max(X_1,X_2,\dots ,X_n)$.

Here are simulated distributions of $T_1$ and $T_2$ in the case $N=300$ and $n=30$, based on 5000 repetitions.

def simulate_T1_T2(N, n):
    """Returns one pair of simulated values of T_1 and T_2
    based on the same simple random sample"""
    tanks = np.arange(1, N+1)
    sample = np.random.choice(tanks, size=n, replace=False)
    t1 = 2*np.mean(sample) - 1
    t2 = max(sample)*(n+1)/n - 1
    return [t1, t2]

def compare_T1_T2(N, n, repetitions):
    """Returns a table of simulated values of T_1 and T_2, 
    with the number of rows = repetitions
    and each row containing the two estimates based on the same simple random sample"""
    tbl = Table(['T_1 = 2*Mean-1', 'T_2 = Augmented Max'])
    for i in range(repetitions):
        tbl.append(simulate_T1_T2(N, n))
    return tbl

N = 300
n = 30
repetitions = 5000
comparison = compare_T1_T2(N, n, 5000)   
comparison.hist(bins=np.arange(N/2, 3*N/2))
plt.title('$N =$'+str(N)+', $n =$'+str(n)+' ('+str(repetitions)+' repetitions)');

We know that both estimators are unbiased: $E(T_1)=N=E(T_2)$. But is clear from the simulation that $SD(T_1)>SD(T_2)$ and hence $T_2$ is a better estimator than $T_1$.

The empirical values of the two means and standard deviations based on this simulation are calculated below.

t1 = comparison.column(0)
np.mean(t1), np.std(t1)

(300.07926666666668, 30.068877736808055)

t2 = comparison.column(1)
np.mean(t2), np.std(t2)

(299.98106666666666, 9.1113762209668376)

These standard deviations are calculated based on empirical data given a specified value of the parameter $N=300$ and a specified sample size $n=30$. In the next chapter we will develop properties of the SD that will allow us to obtain algebraic expressions for $SD(T_1)$ and $SD(T_2)$ for all $N$ and $n$.