A pentagonal bipyramid would work fine. The problem is that reading the result would be difficult. Dice roll on a surface and land on one of the faces, then you read the result (usually) on the face ...

Write out the partial sums of the series into a telescopic sum: $$s_n = \sum_{i=1}^n \frac{1}{i} - \frac{1}{i+1}$$ from here we already see that $$s_n = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} +... View answer Accepted answer 6 votes I'm attempting to solve for one of the beavers, just to see if I've figured it out :) All in all, it is a graph traversal problem which you can solve with backtracking. I believe that he problem is ... View answer 5 votes Here is my attempt, it builds on Cauchy's integral theorem and requires complex numbers, but it works :) Let$$ f(y,x) = \frac{1}{\pi}\frac{\mathrm{e}^{-\frac{|y|}{2}}\mathrm e^{y \frac{i x}{2}}}{1 + ...

The commutator of two transformations $A$ and $B$ represents the difference between applying first $A$ then $B$ and first $B$ then $A$. In linear vector spaces it's pretty easy to follow. Imagine ...

The null space is the space of vectors (of the domain) that are mapped by the operator $T$ into the zero vector (of the codomain) . So, by definition, the null space has to be a subset of the domain ...

The concept of filtration is required to give a formal definition of conditional expectation. In particular, conditional expectation is a random variable because of the sigma algebra of the ...

That's because it depends a bit on which Newton method you refer to. In the one case, it's Newton's root-finding algorithm applied to the gradient of the function: this method will find a local ...

The outer product $u v^T$ more often than not appears as an operator; for instance, as Botond pointed out, if $u$ is a unit vector, then $P_u = uu^T$ is the projector onto the $u$. There are, however, ...

Hint: $$\frac{x^4+ y^4}{x^2+y^2} \leq \frac{x^4 +2x^2y^2+ y^4}{x^2+y^2} = x^2 + y^2.$$

Your expression of the p.d.f. is not very convenient for the estimation. It's better to use $$f_\theta(x) = \theta^{I(0 \leq x \leq 1)}(1-\theta)^{I( 1 < x \leq 2)}.$$ If you spend two minutes ...

Let $\Omega_1 = \{ \omega | X(\omega) = Y(\omega) - Z(\omega)\}$ and $\Omega_2$ be the complement of $\Omega_1$ in $\Omega$. Then $\mathbf P(\Omega_1) = 1$ and $\mathbf P(\Omega_2) = 0$. Then we have ...
I assume you refer to dynamical systems; that is, differential equations of the form $$\dot x = f(x,t,u).$$ These are classified according to which terms appear in $f(x,t,u)$: Time invariant if $f(... View answer 3 votes In the real world, saying that you have a special talent is equivalent to stating, in math, that you have a result. You need a proof. Flawed as it may be, a standardized university curriculum is proof ... View answer Accepted answer 3 votes Yes you can. One way is to rewrite$Y$as the product $$Y = (I_k \otimes x') \begin{bmatrix} A_1 \\ \vdots \\ A_k\end{bmatrix}$$ Then you can use norm inequalities: $$\|Y\| = \left\|(I_k \otimes x') ... View answer 3 votes \Use the replacement \log x = t, then x = \mathrm{e}^t and dt = dx/x, so your integral becomes$$ \int_{-\infty}^\infty e^{r t} \frac{e^{-t^2/2}}{\sqrt{2\pi}} dt $$In which you should ... View answer Accepted answer 3 votes The problem is not in A and B, but in their product AB: the product is not symmetric; hence, there is no clear definition of positive definiteness. In standard parlance, a Hermitian (or ... View answer Accepted answer 2 votes The median is the value such that half of the density of above and half is below it. To find the empirical median you put the data in a long vector. Then you sort the vector. The element in the ... View answer 2 votes The stochastic differential equation$$ \mathrm d X_t = \mu(X_t) \mathrm d t + \sigma(X_t) \mathrm d B_t,$$is just an informal way to represent the integral equation$$ X_{t+h} - X_t = \int_t^{t+h} \... View answer Accepted answer 2 votes Let$f(x,y)$be the mass distribution of the object (the density of the object at point$x,y$). Change the coordinates to polar coordinates: $$g(\rho,\theta) = \rho f(\rho \cos\theta, \rho \sin\theta)... View answer Accepted answer 2 votes The equation of a plane passing through a point (x_0,y_0,z_0) is$$ a(x-x_0) + b (y-y_0) + c(z-z_0) = 0.$$Different values of the coefficients a, b, and c give different planes. Pick any ... View answer Accepted answer 2 votes No it is not necessary that g is invertible. See page 320 of Casella and Berger. The proof of the property is, however, much simpler when g is invertible. View answer Accepted answer 2 votes In my opinion, the events should be X: The number of white balls drawn from Urn 2; the three possible outcomes 2,1, and 0 define the partition you need. B: The ball drawn from Urn 1 is white; ... View answer 2 votes What you have there is a mixture of two exponential distributions. Write the density as$$f(x) = \frac{5}{6}\frac{5}{2}\mathrm e^{-5x/2} + \frac{1}{6}\frac{1}{3}\mathrm e^{-x/3}.$$This describes a ... View answer Accepted answer 2 votes Each sample is generated according to z_i^T H = y_i. If you have N data points (each one consisting of a three-dimensional vector z_i and an observation y_i), you collect them in an N\times 3... View answer Accepted answer 2 votes On this network, variable elimination will not result in a lot of simplification because of the connected structure (that is, not many variables will disappear when we do the elimination). We ... View answer Accepted answer 2 votes You need to marginalize all the variables from the joint distribution:$$p(a,d,f,g) = \sum_{b,c}p(a,b,c,d,f,g) = \sum_{b,c} p(a)p(b|a)p(d|b)p(f|b,c)p(g|c).$$You plug in +d, +f, and \neg g and ... View answer Accepted answer 2 votes The Bayesian network tells you that the joint distribution factorizes according to$$P(a,b,c,d,f,g) = P(a)P(b|a)P(c|a) P(d|b)P(f|b,c) P(g|c),$$So you can compute any truth assignments using the ... View answer Accepted answer 2 votes Judging from the text around your expression, it seems plausible that$T^\star$is the smallest$T$such that$\theta_T \geq E_0[T]$. These are not random variables because they are supposed to be ... View answer 2 votes If the function in continuous, then you can use a lower-bounding argument as follows: If$f(x) > 0$, there is a small enough$\epsilon>0$such that$f(x)>\epsilon$for all$x\in [a,b]\$, You ...