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dolphi86 [110]
3 years ago
11

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).
Mathematics
1 answer:
Georgia [21]3 years ago
8 0

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)

Proof

Let z_a 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.

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