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0

We have $$C(t)=\frac{6}{10}+\frac{e^{it}}{10}+\frac{e^{-it}}{10}+\frac{e^{2it}}{10}+\frac{e^{-2it}}{10}.$$ If $X$ has characteristic function $C$, then the term $e^{it}/10$ comes from the fact that $\mathbb P\{X=1\}=1/10 $, and similarly for the other values. More generally, we can identify characteristic functions which are convex combinations of ...


0

For the case $|cx|<2$, the result follows by showing that the integral is non-negative, as per stocasticboy's comment. To see this, by sketching the graph, it should be obvious that for $|t|<2$, the function $\tfrac 1 t\sin t$ is non-negative (see sketch on Wolfram Alpha).


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It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$. Let $\tau(t):=\inf\{s:A_s>t\}$. Then ...


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If a function is continuous, you can connect its Riemann integral with a limit of a sum of elements. In your case, they are normally distributed. EDIT : let n an integer, and we define the a subdivision of the interval $[0,t]$ , as follows $t_i=\frac{t}{n}i$ with $i \in${ ${0,...,n}$} Since $W$ is a.s continuous, as well as h, we have $t \rightarrow ...


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The exercise can be done without characteristic functions, but the opening poster seems to require to use this way. In the first two cases, we do not have the convergence in distribution of any subsequence to a random variable. Indeed, the value of a characteristic function at $0$ is $1$ and if $t\in (0,2\pi)$, then the characteristic function of $X_n$ and ...


6

$$\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}\varphi_X(t_k-t_{\ell})=E\left(\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}e^{i(t_k-t_{\ell})X}\right)=E\left(\left|\sum_k\xi_ke^{it_kX}\right|^2\right)\geqslant0$$


3

Utterly impossible: $$y?1=y\cdot 1 = y,$$ $$y' = \frac{d(y?1)}{dx}=\frac{dy}{dx}\cdot\frac{d1}{dx} = 0$$


0

With the so-called logarithmic derivative, you can write $$(\log(yz))'=\frac{(yz)'}{yz}=\frac{y'}{y}+\frac{z'}{z}.$$ This does not generalize to the second order.


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It is unclear what you are asking. Why are you defining so many letters? In any case, Convolutions have a property you may be interested in: https://en.wikipedia.org/wiki/Convolution#Integration, but you may need some technical conditions on the function space for these to work. In particular, as the linked page mentions, you'll need Fubini's theorem or one ...


2

My preferred proof is as follows: it suffices to note that for any $\lambda \neq 0$, we have by Sylvester's determinant identity that $$ \det(\lambda I - AB) = \lambda^n\det\left(I - \frac 1{\lambda}AB\right) = \lambda^n\det\left(I - \frac 1{\lambda}BA\right) = \det(\lambda I - BA) $$ Thus, the two polynomials on $\lambda$ are identical for all $\lambda ...


1

There is a mistake in the expression of the characteristic function. We have that $$ \varphi_{X_n}(t)=\operatorname Ee^{itX_n}=\frac1n\sum_{k=1}^ne^{itk/n}. $$ The sum $$ \frac1n\sum_{k=1}^ne^{itk/n} $$ is a right Riemann sum that converges to $$ \int_0^1e^{itx}\mathrm dx $$ as $n\to\infty$.


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It seems you got the idea. Nevertheless: since $a$ is a parameter, you should not use it as an index in the sum and the product. As pointed out by William, you can simplify the expression and get $$\varphi_{X_1+\dots+X_n}(t)=\left(ae^{it}+(1-a)e^{-it}\right)^n.$$


1

I see two errors. $-0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$ and $(-4-\lambda)(-5-\lambda)(-\lambda)-3(-5-\lambda-5)$


2

When you perform Laplacian expansion, remember to switch sign from entry to entry. I think it should be -3 rather than +3 $$0\times \det\begin{bmatrix}1 & 0 \\ 1 & -\lambda\end{bmatrix}+(-4-\lambda)\times \det\begin{bmatrix}-5-\lambda & 0 \\ 5 & -\lambda\end{bmatrix}-3\times \det\begin{bmatrix}-5-\lambda & 1 \\ 5 & 1\end{bmatrix}$$


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I conducted some numerical experimentation, and while I have been unable to provide a rigorous disproof, the results I have obtained seem to suggest that the metrics $d$ and $\rho$ are not strongly equivalent. To see this, let, for each $n\in\mathbb N$, $\mathbb P_n$ denote the probability measure corresponding to the normal distribution with mean $0$ and ...


1

As stated in the question the associated quadratic is $x^2-4x+1=0$ with roots $x=2\pm\sqrt3$, so the general solution to the recurrence is $x_n=A(2+\sqrt3)^n+B(2-\sqrt3)^n$. Putting $x_0=1,x_1=2$ we get $A=B=\frac{1}{2}$. In particular $x_{2015}=\frac{1}{2}\left((2+\sqrt3)^{2015}+(2-\sqrt3)^{2015}\right)$, which we can also write as the nearest integer to ...


0

if $u$ and $v$ are the roots of the characteristic equation, the generating function is of the form $c_1u^n +c_2v^n$. Use $x_0$ and $x_1$ to determine $c_1$ and $c_2$.



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