Darstellung "^3" für den Erwartungswert?

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hey,  wisst ihr vielleicht, wie man auf diese Darstellung "^3" für den Erwartungswert kommt?

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Let $$X_{n} \sim \operatorname{Bernoulli}\left(p_{n}\right)$$, so that $$Y_{n}=X_{n}-p_{n}$$ and $$\sigma_{n}^{2}=p_{n}\left(1-p_{n}\right)$$. As we shall see later (Theorem 4.17), either the Lindeberg Condition (4.8) or the Lyapunov Condition (4.9) will imply that $$\sum \limits_{i=1}^{n} Y_{i} / s_{n} \stackrel{d}{\rightarrow} N(0,1)$$
Let us check the Lyapunov Condition for, say, $$\delta=1$$. First, verify that
$$\mathrm{E}\left|Y_{n}\right|^{3}=p_{n}\left(1-p_{n}\right)^{3}+\left(1-p_{n}\right) p_{n}^{3}=\sigma_{n}^{2}\left[\left(1-p_{n}\right)^{2}-p_{n}^{2}\right] \leq \sigma_{n}^{2}$$
Using this upper bound on $$\mathrm{E}\left|Y_{n}\right|^{3}$$, we obtain $$\sum \limits_{i=1}^{n} \mathrm{E}\left|Y_{i}\right|^{3} \leq s_{n}^{2}$$. Therefore, the Lyapunov condition is satisfied whenever $$s_{n}^{2} / s_{n}^{3} \rightarrow 0$$, which implies $$s_{n} \rightarrow \infty$$. We conclude that the proportion of successes tends to a normal distribution whenever
$$s_{n}^{2}=\sum \limits_{i=1}^{n} p_{n}\left(1-p_{n}\right) \rightarrow \infty,$$
which will be true as long as $$p_{n}\left(1-p_{n}\right)$$ does not tend to 0 too fast.

LG

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