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-1

Denote the mean and the variance of $X_i$ by $m_i$ and $\sigma_i$, respectively. Then, if $n$ is large enough then $$X=\sum_{i=1}^{n}\frac{X_i-m_i}{\sigma_i}$$ can be considered as a random variable of standard gaussian distribution, whose cdf be denoted by $F$. Then $$ P(X<X_0)=E[P(X<X_0|X_0)]=\frac{1}{b_0-a_0}\int_{a_0}^{b_0}F(x)dx.$$ Unfortunately ...


1

Your sequence is going to zero pointwise. On the one hand, $f_k(0) \equiv 0$. On the other hand, if $x>0$ and $k>-\log_2(x)$, then $f_k(x)=0$. Additionally, your sequence is dominated by $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$, which is in $L^p$ for $p \in [1,2)$. So it is definitely going to zero in $L^p$ for $p \in [1,2)$. It does not go to zero ...


2

You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$. For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$. Now, for $L^1$ we can integrate. ...


1

For a fixed $t$, taking logarithm gives $\log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}})$ Using $$1-\cos(x) = \dfrac{x^2}{2} - O(x^4), x\to 0$$ $$\log(1-x) = -x + O(x^2), x\to 0$$ we have $$1- \cos(kt \sqrt{\dfrac{3}{n^3}}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{9k^4t^4}{n^6}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6})$$ and ...


0

Hint: now you want to look at how close $$\cos\left(k t \sqrt{\frac{3}{n^3}}\right) \exp\left(\frac{3 k^2 t^2}{2 n^3}\right)$$ is to $1$.


1

If $\phi(t):=\mathbb Ee^{itX}$ then $\phi(0)=1$. So if $\chi(t)$ denotes the characteristic function up to a constant then: $$\phi(t)=\frac{\chi(t)}{\chi(0)}$$


4

The probability density function is non-negative. Hence, $|f(x)|=f(x)$ and $$ \int |f(x)|\mathrm dx=\int f(x)\mathrm dx=1. $$ It follows that $f\in L^1$.


2

Hint: using the definition of characteristic function, show that if $\phi(t)$ is the characteristic function of a random variable $X$, then the characteristic function of $(X-a)/b$ is $e^{-ita/b} \phi(t/b)$. Also, you have the wrong characteristic function for $X_n$; it's not $e^{eit-1}$ but $e^{n(e^{it}-1)}$.


0

It is sufficient to show that for any $n\in\mathbb{R}^+$ we have: $$ \int_{0}^{1}(1-x)\cos(\pi n x)\,dx = \frac{1-\cos(n\pi)}{\pi n^2}.\tag{1}$$ However, $(1)$ simply follows from integration by parts.


0

The inversion formula states that $f(x) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty}e^{-itx}\phi(t)\,dt$ where $f(x)$ is a probability density function of a random variable $X$ and $\phi(t)$ is the characteristic function of $X$. Since $\phi(t) = \max(1-|t|,0)$ is non-zero only for $-1 < t < 1$, we have: $f(x) = ...


0

It could be nice to use the fact that the Fourier transform of the Fourier transform of a function gives you back the original function (up to constant that depends on the definition of Fourier transform) mirrored in the $y$-axis. Since your functions are even, you can neglect the last part. Just calculate the Fourier transform of $\text{max}\, (1-|t|,0)$. ...


0

Since $a>0$ we have $$\left| e^{r (\imath t-a)} \right| = e^{-r \cdot a} \to 0 \qquad \text{as} \, \, r \to \infty$$ for any $t \in \mathbb{R}$. Therefore, $$\begin{align*} \left[ \frac{e^{x(\imath t-a)}}{\imath \, t-a} \right]_0^{\infty} &= \lim_{r \to \infty} \left[ \frac{e^{x(\imath t-a)}}{\imath \, t-a} \right]_0^{r} \\ &= \frac{1}{\imath ...


0

As yet I've got: $$\frac{1}{2\lambda}\exp\left(-\frac{|x|}{\lambda}\right) = \begin{cases} \frac{1}{2\lambda}\exp\left(-\frac{x}{\lambda}\right)&, x\geq 0 \\ \frac{1}{2\lambda}\exp\left(\frac{x}{\lambda}\right)&,x<0 \end{cases}$$ Now, I define my characteristic function \begin{align*} \varphi_X(t)&=\mathbb{E}(e^{itX}) \\ ...


1

Treat the cases $Z=0$ and $Z=1$ (the only two possible values of $Z$, since $Z$ has Bernoulli distribution).


3

Hint: Let $G$ be a group and $H \leq K$ be two subgroups of $G.$ If $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G.$ EDIT: Let $g \in G$ and consider the automorphism $\phi_g:G \to G, x \mapsto gxg^{-1}.$ Since $K$ is normal, $gKg^{-1} = K.$ So $\phi_g|_K$ is an automorphism of $K.$ Since $H$ is a characteristic subgroup of ...


1

It suffices to consider the case $n=1$ (for $n \geq 2$ use induction). Denote by $\phi$ the characteristic function and $\mu$ the distribution of the random variable $X$. Then $$\begin{align*} \phi(2h)-2\phi(0)+\phi(-2h) &= \int (e^{-\imath \, 2h x} -2 + e^{\imath \, 2hx}) \, \mu(dx) \\ &= 2 \int (\cos(2hx)-1) \mu(dx). ...


1

Suppose $X_n \Rightarrow X$, ie $\mu_{X_n} \Rightarrow \mu_X$. We desire to show that $\phi_{X_n}(\xi) \rightarrow \phi_{X_n}$ as $n \to \infty$ $\forall \xi \in \Bbb R^d.$ Write $\mu_n = \mu_{X_n}$ and $\mu = \mu_X$. Since, by definition, $\mu_n \Rightarrow \mu$, in particular we have that, for each (fixed) $\xi \in \Bbb R^d$, $\mu_n(e^{i x \cdot \xi}) ...


1

$$\left|\frac{1-\cos ht}{t^2}\right|\underbrace{|\varphi_{n_k}(t)|}_{\leq 1}\leq \left|\frac{1-\cos ht}{t^2}\right|\in L^1(-\infty ,\infty )$$ therefore dominated convergence theorem can conclude.


0

Making explicit the my comment above. We know that if $X\sim N(\mu,\sigma)$ then $\phi_X(t)=e^{i\mu t-\frac{1}{2}\sigma^2t^2}$ Then $e^{-x^2}$ is the characteristic function of a $Y\sim N(0,\sqrt{2})$. Notice that it $Z$ is constant $=0$, then $\phi_Z(t)=E[e^{itZ}]=E[1]=1$. The distribution (density!? I don't know well the names in probability theory) of ...



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