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They do achieve the same objective, and we can make the comparison even clearer. Let $\hat F_n$ give the empirical CDF and let $\hat f_n$ be the corresponding pmf that puts a mass of $1/n$ on each observed value (which are almost surely unique here). Then the first moment of $\hat F_n$ is $$ \int x \,\text d \hat F_n = \sum_{x \in \{x_1,\dots,x_n\}} x \hat ...


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The moment-generating function of any random variable $X$ is defined as $$ M(t) := E(e^{tX}).$$ So you can simply take the MGF of whatever geometric distribution you are referring to from its Wikipedia page and evaluate it at $t=1$.


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Let us consider first an example with only $n=3$ random variables. So, we will have $X_{1},X_{2},X_{3}$ are a multiplicative system. And we will prove that $$\mathbb{E}[\prod_{i=1}^{3}(a_{i}X_{i}+b_{i})] =\prod_{i=1}^{3}b_{i}$$ We have the product $(a_{1}X_{1}+b_{1})*(a_{2}X_{2}+b_{2})*(a_{3}X_{3}+b_{3})$, where if we expand it we will have that it is equal ...


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This is an intriguing question because it contains a paradox of intuition as articulated in the question: there are reasons to suggest both that the card player has and does not have an advantage. Let's refine this intuition with a short example. Consider the version of this game where there are $r=b=2$ each of the red and black cards and a player wins only ...


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Let $u$ be the second-highest order statistic, and $v$ the highest order statistic. Then we can write the joint pdf for $u$ and $v$ as the $i=n-1$, $j=n$ case of the theorem 5.4.6 here, $$n(n-1)F(u)^{n-2}f(u)f(v)$$ So we calculate the expectations by integrating this expression times $(v-u)/v$ or $(u/2)(v-u)/v$, integrating over the region with $u<v$. ...


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There is a difference between unconditional mean and conditional mean, as there is between unconditional variance and conditional variance. Mean For a random walk $$ Y_t=Y_{t-1}+\varepsilon_t $$ with $\varepsilon_t\sim i.i.d(0,\sigma_\varepsilon^2)$, the condtional mean is $$ \mathbb{E}(Y_{t+h}|Y_{t})=Y_t $$ for $h>0$. This means that given the last ...


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To see what is happening you need more than one realisation of the random walk, because the mean and variance are summaries of the distribution of the walk, not of any single realisation. This code repeats your code to plot 20 random walks set.seed(1) ys<-replicate(20,{ TT <- 100 y <- ww <- rnorm(n = TT, mean = 0, sd = 1) for (t in 2:TT) { y[...


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In statistical learning theory we pretty much always are considering the joint distribution of $(X,Y)$ and the expected value here is with respect to that distribution, not just the marginal distribution of $X$. In general $$ \text E[L(Y, f(X))] = \int_{\mathcal X \times \mathcal Y} L(y, f(x)) \,\text dP(x, y) $$ where this notation unifies the discrete and ...


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Suppose the fixed effects model is $$y_{ijk}=\mu+\tau_i+\beta_j+\gamma_{ij}+\varepsilon_{ijk}\quad;\small \,i=1,\ldots,a,j=1,\ldots,b,k=1,\ldots,n \tag{$\star$}$$ Here $\mu$ is a general effect; $\tau_i,\beta_j,\gamma_{ij}$ are additional effects subject to $$\sum_i \tau_i=\sum_j \beta_j=\sum_i \gamma_{ij}=\sum_j \gamma_{ij}=0$$ These constraints are ...


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If you want to calculate the expected value analytically, you can use the law of the unconscious statistician. $H\sim N(0,1)$ and $X=f(H)$, then: \begin{align} E(X)=\int_{-\infty}^{\infty}f(h)\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}h^2}\mathrm{d}h \end{align} However, if $f(h)$ is complicated, there may be no analytical solution or it may be very difficult to ...


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This problem is interesting because an intuitive approach leads to a quick solution. Consider a simplified version. Randomly draw every ball from the urn, one at a time, and lay them down in sequence as you do so. The sequence ends with a string of all green or all red balls. You will eventually get to keep this "monochromatic suffix" as a ...


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If you want to calculate the expected value of $x$ numerically, it is quite simple. The result is $$E[x]=\frac{1}{1000}\sum_{i=1}^{1000}f(h_i),$$ where $h_i$ is the $i$-th sample. If you want to calculate it analytically, there is no general way unless you give a specific $f$


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It’s an assumption of his model. Look at the previous page, Eq 5.29. He assumes that $x^{(i)}\sim\mathcal N(\mu,\sigma^2)$, i.i.d. Gaussian. By definition, then $E[x^{(i)}]=\mu$.


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For the mean, you can apply the linearity of expectation a bit more aggressively than you do: since this is just a finite sum we know we can exchange expectation and summation so $$ \text E\left(\sum_{ij} Y_{ij}\right) = \sum_{ij} \text E Y_{ij} = nt\mu $$ and then $\text E \bar Y_{++}$ follows. For the variance, the key step is remembering that the random ...


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You should look into Kaplan Meier analysis, which is a way of handling this type of "right censored" data. In these cases, you want to leverage every data point as much as you can, so you'll still want to use the fact that Person 5's phone hasn't broken in 3 years, even though we don't know how long it will actually last (but we know it's at least ...


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Sadly, there is no closed-form solution for the ATT except in certain cases. The formula for the ATE is the combined coefficient on the A when evaluating the predictors at their means, i.e., -3 + 5*.55 - 10*.3 + 15*(.55*.3) which does equal -.775 as you have figured out. (Note that the final term should include the mean of x1*x2, which in this case happens ...


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There are a couple of issues that can be addressed here. First, In general, the expected value of the product of two random variables does not need to be equal to the product of their expectation. This means that $$ E(XY) \neq E(X)E(Y) $$ The equality only holds when the random variables are independent. But... This links directly to the second issue. Second ...


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Since $\mathbb{E}[X]^2$ is non-negative, then the statement: $$ \text{Var}(X) \leq \mathbb{E}[X^2] $$ Is true in general. I am not sure what you mean by "dependent variables" here since there is only one random variable $X$.


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