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I think you forgot to divide by the determinant of the Fisher Information Matrix. You appear to have multiplied by it (or something).

I think double check your Var, the beta terms should end up cancelling out I think.

$\noindent It is the approximate \textbf{variance} of $\widehat{\beta}$ (for large $n$), yes.$

$\noindent First step would be to find the 1,1 entry of the inverse of the Fisher information matrix (this would be the approximate variance of $\widehat{\alpha}$ for large $n$). To eliminate $\beta$'s from your answer, use the relationship between $\widehat{\alpha}$ and $\beta$ from the MLE...

$\noindent The fact that the terms form an arithmetic progression with \underline{common difference $5$} tells us that the general term is of the form $5k+b$ for some constant $b$ (where $k$ will be our summation index). Then depending on what you want to start your $k$ as, you can find $b$...

By the way, this is a Pareto Distribution with scale parameter 1 (and shape parameter alpha). This distribution is related to the "80-20" law or "Pareto principle", which you may have heard of. A fact about Pareto distributions is that the conditional distribution of a Pareto r.v. X given the...

$\noindent Let $I = \int_{0}^{1}x^{3} \left(x^{4} - 3\right)^{4}\, \mathrm{d}x$. Let $u = x^{4} - 3$, so $\mathrm{d}u = 4x^{3}\, \mathrm{d}x \Rightarrow x^{3}\, \mathrm{d} x = \frac{1}{4}\mathrm{d}u $. \textsl{For the limits, when $x = 0$, $u = 0 - 3 = -3$, and when $x = 1$, $u = 1 - 3 = -2$.}...

$\noindent Let $\omega = \frac{1}{3}$, then by SHM facts, we know $\ddot{x} = -\omega^{2} x$. Thus acceleration $\ddot{x}$ is greatest when $x$ is least. On the given interval, this is actually when (and only when) $t = 0$, since $x(t) \geq 0$ for all $t \in [0,2\pi]$ with equality iff $t = 0$.$...

Not sure if this is the reason but make sure to have spaces between the inequality signs and neighbouring text when typing them on BOS (otherwise it appears to mess up sometimes).

If it's a continuous case (joint PDF), we could integrate the joint PDF over the region in the x-y plane S := {(x,y) : x + y < 1}. (In other words, calculate a double integral.)

It's the slope of the tangent at the given point (and we're given that tangent's equation (we're given that the line is tangent)).

4

It is not. There is an adjustment factor.

$\noindent (This question is showing that if $X$ is log-normal, then $\ln X$ is normal.) Just determine the pdf or cdf of $Y$ and you should be able to see it is the pdf/cdf of a $\mathcal{N}\left(0,\sigma^{2}\right)$ random variable, and hence $Y$ has that distribution. To do this question, you...

$\noindent For (ii), let $B = PJP^{-1}$ where $J$ is the Jordan form of $B$, then use what you know about the eigenvalues of $B$ and the fact that $B + I= P(J+I)P^{-1}$ to deduce the result.$ $\noindent And I think (iii) is simpler than you (probably) think.$

75 divided by 3 is not 28.

It's used in many hypothesis tests, for example.

Almost all the terms cancel. If you can't see it immediately, it might help you to write out the terms of each sum in long notation.

$\noindent It's a \emph{standardised} version of $\overline{X}$. I can't see the context but I'm guessing $\overline{X}$ is the average of $n$ i.i.d. $\mathcal{N}\left(\mu, \sigma^{2}\right)$ random variables. This implies that $\overline{X}$ is a normal random variable with mean $\mu$ and...

It's very close to 0. Here's a numerical calculation via WolframAlpha: http://www.wolframalpha.com/input/?i=Pr(Z+%3E+4.38),+Z+~+N(0,+1) .