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You can decompose the exponential distribution into a sum of two terms, which are not both gamma distributed. Let A,B,ε be independent where A,B are exponentially distributed and ε takes the values 0 …

We can show that $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$ so that the inequality holds with $c=1/2$ and $C=\sqrt{e}$. For $\epsilon\ge1$ the right hand side …

The answer to your amended question is yes. In fact, for any $\epsilon\in[0,1)$ we have $$ \mathbb{P}(\vert S\vert > \epsilon)\ge (1-\epsilon^2)^2/3. $$ So, we can take $\delta = 1-(1-\epsilon^2)^2/3$ …

If $\mathbb{E}\left[\lVert[ X\rVert^d\right]$ is finite then the integral in the question is necessarily finite. As mentioned, this holds whenever $\pi$ is radially decreasing. However, in the general …

You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process Xt − X0 = bt with constant b. In fact, you can't identify it b …

As I mentioned in my comment, you can prove the statement and its converse by looking at the moment generating function. Supposing that f ≥ 0 and λ > 0 is a real number, the following is true for a Po …

Letting $\mu_{a,\Sigma}$ be the Gaussian measure with covariance matrix $\Sigma$ and mean $a$. Then (double) the variation distance can be written as $$ \left\lVert\mu_{a_1,\Sigma_1}-\mu_{a_2,\Sigma_2 …

To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent if and only i …

No, it is not true that a process W satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows. This construction is rather contrived, and I …

No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$. $W$ is a standard Brownian motion. $U$ is an $ …

I was just looking through a book which proves many interesting and rather difficult results on Brownian motion (pdf link, website link), and it seems that the Kolmogorov zero-one law applies to most …

Consider the following continuous Markov process X, starting from position x if x = 0 then Xt = 0 for all times. if x ≠ 0 then X is a standard Brownian motion starting from x. This is not strong M …

There's already been some good responses, but I think it's worth adding a very simple example showing what can go wrong if you don't use Polish spaces. Consider $\mathbb{R}$ under its usual topology, …

This can answered without any complicated maths. It can be related to the following: Imagine you have $N$ marked cards in a pack of $m$ cards and shuffle them randomly. What is the probability that t …

As has been mentioned in previous answers, the moments do not uniquely determine the distributions unless certain conditions are satisfied, such as bounded distributions. One thing you can say, is tha …