4 votes
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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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  • 229
3 votes

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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2 votes
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Expected value of $X^3$

If $X\sim f_X(x)$, then $$ \begin{aligned} \mathbf{E}\left[X\right] &= \int_{-\infty}^{+\infty}xf_X(x)dx \\ \mathbf{E}\left[g(X)\right] &= \int_{-\infty}^{+\infty}g(x)f_X(x)dx, \text{for some ...
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  • 1,285
1 vote

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

We can also calculate $\mathsf E(N)$ directly. We let $\mathbb N=\mathbb Z_{\ge 1}$ and assume that $(X_n)_{n\in\mathbb N}$ is a sequence of iid random variables such that $\mathsf P(X_1 = m) = p (1-p)...
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1 vote
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$\mathbb{E}[Y|X+Y=z]$

Everything checks out save the typographic error: you jumped from an expression in $y$ to an expression in $x$. $X+Y\overset{\small d}{=}\mathcal{P}\left(\lambda_{1}+\lambda_{2}\right)$, so \begin{...
1 vote

Expected value of $X^3$

I know that mistake is somewhere when finding $f_{X^3}(x)$ Take some uniformly spread sample points, say $\{0.00,0.25,0.50,0.75,1.00\}$, and examine their cubes, $\{0, 0.015625, 0.125, 0.421875, 1\}$....
1 vote

Conditional Expectation of multivariable function

Counterexample: $Y=X$ with $X$ symmetric, e.g., $X\sim\cal N(0,1)$. Take $g(x,y):=xy$ and $h(y):\equiv0$. Then $$\Bbb E[g(X,y)]=y\,\Bbb E[X]=0=h(y)$$ for any $y\in\Bbb R$, but $$\Bbb E[g(X,Y)\mid Y=y]=...
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1 vote
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Exercise 1.9: Show that if a random variable X is independent of an event A, then a function of the random variable g(X) is also independent of A

$\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A\sum_{k=1}^n\alpha_k\mathbb{I}_{B_k}(x)d\mu_X(x) = \sum_{k=1}^n\alpha_k\int_A\mathbb{I}_{B_k}(x)d\mu_X(x)$ $ = \sum_{k=1}^n\alpha_k\mathbb{P}\{X\in \...
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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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