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### Calculating $\mathbb{E}[N]$ for $N = \displaystyle \min_{n\in \mathbb{N}}\Big\{\sum_{i=1}^{n}{X_i\geq5000\Big\}}$, using Wald's lemma

I reached the same conclusion as Henry (+1) in a more roundabout manner. What follows is the gist of my thoughts: Recall that a geometric random variable models the number of independent Bernoulli ...
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### Calculating $\mathbb{E}[N]$ for $N = \displaystyle \min_{n\in \mathbb{N}}\Big\{\sum_{i=1}^{n}{X_i\geq5000\Big\}}$, using Wald's lemma

Geometric distributions have the memorylessness property, which makes this much easier if you regard it as a sequence of attempts stopping with the first success at $5000$ or more attempts. You can ...
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$\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A\sum_{k=1}^n\alpha_k\mathbb{I}_{B_k}(x)d\mu_X(x) = \sum_{k=1}^n\alpha_k\int_A\mathbb{I}_{B_k}(x)d\mu_X(x)$ $= \sum_{k=1}^n\alpha_k\mathbb{P}\{X\in \... • 40.3k 1 vote ### Striking applications of linearity of expectation It also helps calculate the falling factorial momentum of the Poisson distribution. As recall: The falling factorial momentum is defined as:$\$\mathbb E[(X)_r]=\mathbb E[X(X-1)\cdots(X-r+1)]=r!\...
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