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.
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 = \,
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