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Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions. In fact, the PDF is $$f(y;p)=2p^2y^{-3}\mathbf 1_{y\geqslant p}$$ hence, th …

Hint: The delta-method applied to $\sum\limits_{k=1}^nx_k^2$ yields $$\sqrt{n}(\hat\theta_n-\theta)\to N(0,\tfrac23).$$ To prove this, one starts with the CLT expansion $$\sum\limits_{k=1}^nx_k …

The likelihood based on the sample $(x_k)$ is by definition $L(\theta)=f_\theta(x_1)f_\theta(x_2)\cdots f_\theta(x_n)$. In the present case, $L(\theta)=\theta^{-n}\mathbf 1_{x_1\leqslant\theta}\mathbf …

If $(Y_k)$ is i.i.d. standard exponential, that is, with PDF $$f(y)=e^{-y}\mathbf 1_{y>0}$$ then, for every $n\geqslant1$, $T_n=Y_1+\cdots+Y_n$ has PDF $$f_n(t)=\frac{t^{n-1}}{(n-1)!}e^{-t}\mathbf 1_{ …

The shortest route and the most rigorous one are the same, which simply require to write correctly each PDF involved, including the support conditions in the PDFs, as they should be. Here, for e …

Consider a positive integer $n$ and a set of positive real numbers $\mathbf p=(p_x)$ such that $\sum\limits_xp_x=1$. The multinomial distribution with parameters $n$ and $\mathbf p$ is the distributio …