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Let $X=(X1,…,Xn)$ and $x=(x1,…,xn)$. To find $\hat{\theta }_{ML}$, we first find $f_X(x;θ)$. Since the $W_i$'s are independent, so are the $X_i$'s. Hence,$$\displaystyle f_ X(x;\theta )\displaystyle = \prod _{i=1}^{n}f_{X_ i}(x_ i;\theta )\displaystyle = \begin{cases} \displaystyle \prod _{i=1}^{n}e^{-(x_ i-\theta )}, & \mbox{if } x_ i \geq \theta ~ ...


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We want to obtain the distribution of the estimator $\hat{\theta} = max_{i = 1}^n{X_i}$: $$P(max_{i=1}^n{X_i} \leq x) = \prod_{i=1}^nP(X_i \leq x) = P(X_1 \leq x)^n = (\frac{x - 0}{\theta - 0})^n = \frac{x^n}{\theta^n} I_{[0, \theta]}(x)$$ Which means that, differentiating the function with respect to $x$, we obtain: $$P(max_{i_1}^n{X_i} = x) = ...


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There is nothing wrong with the derivation, only problem with experimental setup. The experiment is calculating expected value of standard deviation and not of variance. Change RMSExptAvg(S, n) = mean([RMSExpt(S, n) for i in 1:10000])) to this: RMSExptAvg(S, n) = sqrt(mean([RMSExpt(S, n)^2 for i in 1:10000])) Then equations match up exactly. Getting ...


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Let $Y_n$ be $+1$ if $X_n$ is $6$, $-1$ if $X_n$ is $1$ and $0$ in all other cases. Then \begin{align*} E(Y_n)&=1\left(\frac{1}{6}+e\right)+(-1)\left(\frac{1}{6}-e\right)=2e,\\ E(Y_n^2)&=1^2\left(\frac{1}{6}+e\right)+(-1)^2\left(\frac{1}{6}-e\right)=\frac{1}{3}, \end{align*} and $V(Y_n)=\frac{1}{3}-4e^2$. Then, the estimator ...



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