Question

Let X1, . . . ,Xn ∈ R be independent random variables with a common CDF F0. Let Fn be their ECDF and let F be any CDF. If F = Fn, then L(F) < L(Fn).

Answer 1

Answer:

See the proof below.

Step-by-step explanation:

For this case we need to proof that: Let $X_1, X_2, ...X_n \in R$ be independent random variables with a common CDF $F_0$. Let $F_n$ be their ECDF and let F any CDF. If $F \neq F_n$ then $L(F) <L(F_n)$

Proof

Let $z_a<z_2<....<z_m$ different values in the set {$X_1,X_2,...,X_n}$} and we can assume that $n_j \geq 1$ represent the number of $X_i$ that are equal to $z_j$.

We can define $p_j = F(z_j) +F(z_j-)$ and assuming the probability $\hat p_j = \frac{n_j}{n}$.

For the case when $p_j =0$ for any $j=1,....,m$ then we have that the $L(F) =0< L(F_n)$

And for the case when all $p_j >0$ and for at least one $p_j \neq \hat p_j$ we know that $log(x) \leq x-1$ for all the possible values $x>0$. So then we can define the following ratio like this:

$log (\frac{L(F)}{L(F_n)}) = \sum_{j=1}^m n_j log (\frac{p_j}{\hat p_j})$

$log (\frac{L(F)}{L(F_n)}) = n \sum_{j=1}^m \hat p_j log(\frac{p_j}{\hat p_j})$

$log (\frac{L(F)}{L(F_n)}) < n\sum_{j=1}^m \hat p_j (\frac{p_j}{\hat p_j} -1)$

So then we have that:

$log (\frac{L(F)}{L(F_n)}) \leq 0$

And the log for a number is 0 or negative when the number is between 0 and 1, so then on this case we can ensure that $L(F) \leq L(F_n)$

And with that we complete the proof.