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$\noindent In general for a random variable distributed according to an exponential distribution with rate parameter $\lambda$, the mean is $\lambda^{-1}$ and the variance is $\lambda^{-2}$. If you know these facts, then you can reduce that the second moment ($\mathbb{E}\left[X^2\right]$) is...

Note that to get E[T^2], we're integrating x^2 times the PDF, not x times the PDF.

$\noindent I'll just give an example here. Say $X \sim \mathrm{Exp}\left(4\right)$, so has PDF $f_{X}(x) = 4e^{-4x}$ ($x \geq 0$). Then for example, $\mathbb{E}\left[X^2 + 3X -7\right] = \int _{\text{range of }X} \left(x^2 + 3x -7\right)f_{X}(x)\, \mathrm{d}x = \int_{0}^{\infty}\left(x^{2} + 3x...

That's an unnecessarily tedious way to do it if you just want the variance of Y, but yeah, to find the expectation of that quadratic quantity in X, if we're given the PDF or PMF, we can easily find it using the Law of the Unconscious Statistician (LOTUS). I can demonstrate the integral/sum...

$\noindent Yeah we can. Recall the property of variance $\mathrm{Var}\left(aX + b\right) = a^2 \, \mathrm{Var}(X)$, for any constants $a,b$. So if we're given the distribution of $X$ in your example, we can calculate its variance and use that to find $Y$'s variance, as we will have $Y$'s...

To both: in general no.

Re: Discrete Maths Sem 2 2016 I guess it's basically looking into the heart of the question. (It's phrased in terms of balls and boxes, but it's essentially a question about how many surjections we can make on given-sized (finite) domains and codomains.)

Not really. That's just the probability density function (PDF) of the normal distribution. All continuous distributions have their PDF that characterises them. (For discrete distributions, the analogue to the PDF is often called a 'probability mass function' (PMF).)

$\noindent We can approximate some distributions with the Normal distribution sometimes (using Central Limit Theorem). E.g. we can do a normal approximation to the Binomial distribution. If $X \sim \mathrm{Bin}(n,p)$ and $n$ is large, then approximately, $X$ is normal with mean $np$ and variance...

Pretty much, the main thing to focus on in terms of how to get that matrix is the columns (as these came from the procedure described above).

$\noindent It's because the map $T$ is the transpose map. Remember, the matrix of a linear map $T$ with respect to a basis $\mathcal{B} = \left\{\vec{b}_1, \vec{b}_{2},\ldots , \vec{b}_n\right\}$ in the domain and $\mathcal{C}$ in the codomain (for our case $\mathcal{B}$ and $\mathcal{C}$ were...

(I was being lazy in some places when writing "solution" above.)

$\noindent Note that clearly there are solutions if $n=1$ or $2$. To get solutions for any odd power $2k+1$, start with a known solution $x^2 + y^2 = z$, then multiply both sides by $z^{2k}$, so we have $\left(xz^{k}\right)^{2} + \left(yz^{k}\right)^2 = z^{2k+1}$ as our solution. The idea for...

Oh yeah sorry, I got a bit complacent/was on autopilot and miscalculated the transposes of P and Q. It should be T(P) = P and T(S) = S, of course, and those answers are right. Thanks!

$\noindent b) For simplicity we'll refer to the basis matrices as $P, Q, R, S$ in that order. So just transpose them to find that $T(P) = P, T(Q) = R, T(R) = Q$ and $T(S) = S$. It follows that the desired matrix is: $B = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &...

Which parts do you want help with? All of them? What progress have you made so far? Here's how to do the first part. $\noindent The result (namely that any eigenvalue of $T$ must be $\pm 1$) actually will hold for any linear map with $T\circ T = I$, where $I$ is the identity map (i.e. any...

$\noindent a) Suppose $\lambda$ is an eigenvalue for $P$ with eigenvector $\mathbf{v}$, so $P(\mathbf{v}) = \lambda \mathbf{v}$. Apply $P$ to both sides: $P(P(\mathbf{v})) = P(\lambda \mathbf{v}) \Rightarrow P(\mathbf{v}) = \lambda P( \mathbf{v} ) \Rightarrow \lambda \mathbf{v} = \lambda^2...

Re: HSC 4U Integration Marathon 2017 $\noindent Let $a$ be a real constant with $|a| \neq 1$. Find $\int \frac{1-a^{2}}{1-2a\cos x + a^{2}}\, \mathrm{d}x$.$

$\noindent Yep. Recall the general identity $\det \left(cM\right) = c^{n}\det M$, if $M$ is an $n\times n$ matrix and $c$ a scalar.$

$\noindent 1) Yeah that's essentially what the $P_{2}\to P_{3}$ would mean (though technically it means polynomials of degree \emph{at-most} $2$ get inputted and \emph{at-most} $3$ get outputted. We can still input linear polynomials and lower degree in addition to quadratics, and we can get out...