7

One way to make your mgf approach to the problem easier is to use the power series $$(1-t^2)^{-1}=\sum_{j=0}^{\infty} t^{2j}$$ Differentiating the rhs repeatedly is much easier than differentiating the lhs. (note this only applies for $|t|<1$ but as you are differentiating at $t=0$ it still works). You should see that it will just be a factorial. ...


6

If you don't care to do unnecessary calculation, it is convenient to view your distribution as an equal mixture of an Exponential and its negative: $$\frac{1}{2} e^{-|x|} = \frac{1}{2} e^{-x}\,\mathcal{I}(x\gt 0) + \frac{1}{2} e^{x}\,\mathcal{I}(x \lt 0).$$ Because $((-1)^n + (1)^n)/2$ is either $-1+1=0$ or $(1+1)/2 = 2/2 = 1$ as $n$ is odd or even, ...


6

Since the OP seems to be having difficulty with the various hints in the comments and the other answers, here is a heuristic method that yields the right answer in this instance. \begin{align} E[\exp(tX)] &= E\left[1 + tX + \frac{(tX)^2}{2!} + \frac{(tX)^3}{3!} + \cdots\right]\\ &= 1 + tE[X] + \frac{t^2}{2!}E[X^2] + \frac{t^3}{3!}E[X^3] + \cdots\tag{...


6

Hint: This is an example of a probability density function that is symmetric about zero: $$f_X(0+x) = f_X(0-x) \quad \quad \quad \text{for all } x \in \mathbb{R}.$$ Visually, this means that the distribution is reflected around the zero line, and is the same on both sides. See if you can use this property to figure out (and then prove) what the odd ...


3

The day three The elders of the statistics guild have discovered a problem in the divine parameters. There is no single solution possible because the system is over-determined. We can scale the different values of the groups and the results will remain true. For instance when we divide the 'constant' coefficient by two and at the same time multiply the '...


3

We can pose PCA as a variance maximization problem. These are some of the hints: The objective is to find the directions in which the variance, $\Bbb E(\vec X \vec X^T)$, is maximum. Let $\vec w$ denote the unit vector direction along which the variance is maximum. The variance along this direction is given by: \begin{aligned} \sigma_{\vec{\omega}^{2}}^{...


1

The integral is actually the following: $$p(\mathbf{y}|\mathbf{x})=\int p(\mathbf{y|\mathbf{x},\mathbf{w}})p(\mathbf{w})d\mathbf{w}$$ Here, $p(\mathbf{y|\mathbf{x},\mathbf{w}})$ is the distribution of $\mathbf{y}$ when we know $\mathbf{x}$ and $\mathbf{w}$, in which the relation is actually deterministic, i.e. $\mathbf{y}=\mathbf{\Phi(x)}\mathbf{w}$. So, $\...


1

See @Glen-b's comment. What follows finds the MGF for an ordinary geometric distribution (where $\alpha = 1-\theta)$ that counts the number of failures before the first success, and so has support on the nonnegative integers. [There are other versions of the geometric distribution.] To make typing easier, I will let $p$ be the success probability and $q = ...


1

So I calculated the first, second, third, and fourth derivatives. I got $E(X^1)=0$, $E(X^2)=2$, $E(X^3)=0$, and $E(X^4)=12$. These derivatives are quite long to compute at this point, so I m wondering if there is an easier way to go about this to obtain a formula for the evens. You could use the Taylor series expansion: $$\frac{1}{1-t^2} = \sum_{k=0}^\...


1

No going back case only I'm addressing the zero-length jumps case only, i.e. no going back and the frog is allowed to remain still at a given step. Without considering a clock-like device and assuming that remaining still in one clock tick counts as one jump means just considering the other's puzzle conditions. It does not have to be a precise clock or ...


1

The answer by Glen_b frames the HPD as a set of simultaneous equations that can be solved via numerical methods. This is one possible way to compute the HPD. An alternative method is to frame the HPD as an optimisation problem, and solve this via numerical methods. Computing the HDR via optimisation: Suppose you have a random variable $X \sim f$ where the ...


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