Let X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathbf{P} for some distribution \, \mathbf{P}\,, and let F denote its cdf. Let F_ n denote the empirical cdf. Then it holds for every t \in \mathbb {R} that
\sqrt{n} (F_ n(t) - F(t)) \xrightarrow [n \to \infty ]{(d)} \mathcal{N}(0, \sigma ^2)
for any fixed t and for some asymptotic variance \sigma ^2. What is the asymptotic variance \sigma ^2 of F_ n(0) in terms of the values of the cdf F? (Enter F(x) for F(x) for any numerical value x.)
\, \sigma ^2 = \,
To find the asymptotic variance σ^2 of F_n(0), we need to consider the properties of the empirical cdf F_n.
The empirical cdf F_n is defined as the fraction of observations that are less than or equal to t:
F_n(t) = \frac{1}{n}\sum_{i=1}^{n} I(X_i \leq t),
where I(X_i \leq t) is the indicator function taking the value 1 if X_i \leq t and 0 otherwise.
At t = 0, the empirical cdf F_n(0) represents the fraction of observations that are less than or equal to 0:
F_n(0) = \frac{1}{n}\sum_{i=1}^{n} I(X_i \leq 0).
To find the asymptotic variance σ^2, we need to consider the distribution of \sqrt{n} (F_n(0) - F(0)) as n tends to infinity.
Using the Central Limit Theorem, we know that \sqrt{n} (F_n(t) - F(t)) converges in distribution to a standard normal distribution:
\sqrt{n} (F_n(t) - F(t)) \xrightarrow [n \to \infty ]{(d)} \mathcal{N}(0, 1),
where \mathcal{N}(0, 1) represents a standard normal distribution.
In our case, we are interested in t = 0:
\sqrt{n} (F_n(0) - F(0)) \xrightarrow [n \to \infty ]{(d)} \mathcal{N}(0, 1).
Therefore, the asymptotic variance σ^2 of F_n(0) is 1.
To find the asymptotic variance \(\sigma^2\) of \(F_n(0)\), we can use the Delta Method.
The Delta Method states that if \(\sqrt{n}(Y_n - \theta) \xrightarrow[]{(d)} \mathcal{N}(0, \sigma^2)\), and \(g\) is a differentiable function, then \(\sqrt{n}(g(Y_n) - g(\theta)) \xrightarrow[]{(d)} \mathcal{N}(0, \sigma^2[g'(\theta)]^2)\).
In this case, \(Y_n = F_n(0)\) and \(\theta = F(0)\). We need to find the derivative of \(F\) at 0, denoted as \(F'(0)\), to calculate the asymptotic variance.
Taking the derivative of \(F\) with respect to \(t\), we get:
\(f(t) = \frac{d}{dt} F(t)\)
Now, let's substitute \(t = 0\) in the above equation:
\(f(0) = \frac{d}{dt} F(0)\)
The derivative of the cdf at \(t = 0\) gives us the value of the pdf at \(t = 0\). Let's denote this value as \(f(0)\).
Therefore, \(f(0) = F'(0)\).
Finally, using the Delta Method, the asymptotic variance \(\sigma^2\) of \(F_n(0)\) is given by:
\[\sigma^2 = [f(0)]^2\]
So, the asymptotic variance \(\sigma^2\) of \(F_n(0)\) can be obtained by evaluating \([f(0)]^2\), where \(f(0) = F'(0)\).