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As you already know, considering the function $$G_n(x)=\log\left(n+\frac{n-1}{x-1}\right),$$ defined on $(1,+\infty)$, $\hat\lambda_n$ solves the identity $$\hat\lambda_n M_n=G_n(\hat\lambda_n M_n).$$ As you noted, this identity has no analytical solution. However, the function $G_n$ decreases on $(1,+\infty)$ from $G_n(1)=+\infty$ to $G_n(+\infty)=\log(n)$ ...

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Fix $1 \leq l \leq k$, and $t_0 \in[0, 1]$ as in the assumption. Now, choose $0 \leq j_0 \leq n - 1$ such that $\left|\frac{j_0 + 1}{n} - t_0\right| \leq \frac{1}{n}$, then we have \begin{align*} \left|p(\frac{j_0 + l + 1}{n}) - p(\frac{j_0 + l}{n})\right| &\leq \left|p(\frac{j_0 + l + 1}{n}) - p(t_0)\right| + \left|p(\frac{j_0 + l}{n}) - p(t_0)\right| ...

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Let's assume: Gaussianized data has a normal distribution with zero mean and unit variance The output layer uses a softmax layer The cost function is cross entropy which measures how much the distribution of the output layer is different from the desired output distribution Now if the desired output is a sample from Gaussian distribution and you run the ...

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Note the Following. Let $\bar{e}$ denote the mean of e. Then I can re-write your model as $y_i = (\beta_1 + \bar{e}) + \beta_2x_i + (e_i - \bar{e})$. This model is identical to yours and now has a mean-zero error term (though, as you point out, it is still not normally distributed), but the intercept will be "biased" by the mean of the original error.

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Not yet a solution: For $f(\mathbf{y})=a(\mathbf{y})-b(\mathbf{y})$ one can take another random variable $f^{'}(\mathbf{y})=a(\mathbf{y})+b(\mathbf{y})$ Since expectation is a monotone operation $E[|f(\mathbf{y})|]\leq E[f^{'}(\mathbf{y})]$, as $|f|\leq f^{'}$. Now it suffices to show that $E[f^{'}(\mathbf{y})]$ has a finite mean. As expectation is linear ...

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Large deviation gives you an estimate of the probabilities in the non-typical regime, whereas CLT gives you an estimate of the probabilities in the typical regime. Suppose $X_i$ have finite variance and are a.s. positive. Then CLT gives, $$|P(\frac{\sum_i X_i - n\mu}{\sigma \sqrt{n}} \ge x) - \Phi(-x)| \to 0.$$ But CLT does'nt say anything if you let $x$ ...

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With a finite variance, you have the Central Limit Theorem which broadly speaking tells you that for large $n$ the mean is approximately normally distributed with mean $\mu$ and variance $\frac{\sigma^2}{n}$. So for large $n$ $$P(|\bar{X}_n - \mu| \geq \epsilon) \approx 2\Phi\left(-\frac{\epsilon \sqrt{n}}{\sigma}\right)$$ As an illustration, with ...

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You may have a look at the large deviations theory. In this case, when the distribution is regular enough, the Cramer theorem states: $$\frac 1n\log P\left(\frac 1n [X_1 + \dots + X_n] > \mu + \epsilon\right) \to -\left[\sup_{t\in \Bbb R} (\mu + \epsilon)t - \log Ee^{tX} \right]$$ The condition being in that case that $Ee^{tX}$ exists. So the good ...

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