### Math proof required (Probability)

FromPr[E] = Pr[E|A].Pr[A] + Pr[E|A'].Pr[A']how can we provePr[E] <= Pr[E|A] + Pr[A']...Read more

FromPr[E] = Pr[E|A].Pr[A] + Pr[E|A'].Pr[A']how can we provePr[E] <= Pr[E|A] + Pr[A']...Read more

Before asking my question I must first explain the context in which it is formulated.Let $M\subset \mathbb{R}$, if $M$ is not countable then we have $M$ with Borel $\sigma$-algebra denoted by $\mathcal{B}$. If $M$ is countable then we have $M$ with discrete $\sigma$-algebra (i.e. iduced by $\mathcal{P}(M)$). Definition: A random recurrent sequence over a set $M$ is a sequence of random variables $(X_{n})$ with values in $M$ (i.e, $X_{n}(\omega)\in M$) defined over a probability space $\left(\Omega,\mathcal{F}, \mathsf{P} \right)$ such that $(...Read more

Let $X, X'$ be independent with $X \sim p(x)$, $X' \sim r(x)$ for $x, x' \in X$.I don't understand this equation: $\sum p(x)r(x)=Pr(X=X')$What is intuitive to me is if $X \sim p(x)$, $X' \sim p(x)$ for $x, x' \in X$, then $\sum p(x)p(x) = Pr(X=X')$. Could anyone please explain a bit about why $\sum p(x)r(x)=Pr(X=X')$ above? Does it mean that $\sum p(x)r(x) = \sum p(x) p(x)=Pr(X=X')$ ?Source: http://staff.ustc.edu.cn/~cgong821/Wiley.Interscience.Elements.of.Information.Theory.Jul.2006.eBook-DDU.pdf#page=67...Read more

I see that all my understanding of statistics (& so of probability which is a branch of statistics) from high school came from the mathematics textbook and it all appears too mathematical to be accepted as a branch of mathematics but why then it isn't considered a branch of mathematics?Edit 1The first two answers I've got are contradicting each other one is claiming that it (statistic) falls under the domain of measure theory which is a branch of mathematics. So, it's entirely mathematical. The other one is saying that they are different....Read more

Let $X$ be a random variable with probability distribution $f_X$ Then $f_X$ is a function, $f_X : A \to B, c \mapsto d$What exactly are $A,B,c,d$?I am sorry if this question is basic, as I am from a computer science background and we do not usually specify these functions rigorously. In fact I am reading a paper right now and it says "let $p$,$q$, be probability distributions that generates data $X,Y$" - not very rigorous I tell you....Read more

If I roll a six sided dice twice there is a $1$ in $6$ chance that the results will sum up to $7$ (giving an average of $3.5$ per dice). And if I only roll it once it is not possible to get a $3.5$.The more I roll it the more likely that the average result will be close to $3.5$.However for $n$ rolls what is the probability that the average result is exactly $3.5$, and how can I generalize this to an $m$ sided dice?...Read more

Once upon a time I read about nontransitive dice - sets of dice where "is more likely to roll a higher number than" is not a transitive relation. After the surprise wore off, I wondered - just how far can this phenomenon be pushed? The linked Wikipedia page has a section called "Freivald's investigation" which states that if $n$ dice are arranged in a circle each with probability $p$ of being greater than the next in line, then $p<3/4$ (with $p$ otherwise allowed arbitrarily close). However, it has the infamous tag of [citation needed] and I...Read more

Hi guys I am having problems deriving $P(X = k)$ if $P(X = k|X+Y = n)$ = ${n}\choose{k}$ $\times$ $2^{-n} $X and Y are i.i.d. random variables with values in $\mathbb{N_0}$.After playing a bit with the formula and using independency of the variables I get to:$P(X = k)$ = ${n}\choose{k}$$\times$ $2^{-n}$$\times$ $\frac{P(X+Y = n)}{P(Y = n-k)}$and I could theoretically rewrite $ P(X+Y = n)$ = $ \sum_{k = 0}^{n}P(X = k) P(Y = n-k)$, although I am not sure whether the indexing is correct and it does not really help me to get further.I know from an ...Read more

I was looking at a proof of the string law of large numbers, and am having trouble finding where the proof uses the assumption that the random variables are identically distributed. I'll reproduce the proof below, but the link can be found here (https://www.math.ucdavis.edu/~tracy/courses/math135A/UsefullCourseMaterial/lawLargeNo.pdf). To start, we have Chebyshev's inequality and the easy side of Borel-Cantelli (both of which I understand).I'll save time by assuming the independent real-valued random variables $(X_{k})_{k \in \mathbb{N}}$ on pr...Read more

Problem: Let $X_1$ and $X_2$ be independent and uniformly distributed random variables over (0,2). Determine the Density Function of $(X_1/X_2)$.Here are my thoughts on the problem:$\vec{x}=(x_1,x_2)$:$(0,2)\times(0,2)$. So $f_1(x_1)=2, f_2(x_2)=f_1(x_2)=2$ for $\forall x_1,x_2\in(0,2)$ such that $f(x_1,x_2)=\frac{f_1(x_1)}{f_1(x_2)}=1$$y_1=\frac{x_1}{x_2}$ and $y_2=x_2\Rightarrow x_2y_1=x_1, y_2=x_2$. Let $v_1=(x_2y_1)$ and $v_2=y_2$. Using $g(y_1,y_2)=f(v(y_1,y_2))|J_v(y_1,y_2)|$, (where $J_v$ is the $2\times2$ Jacobian Matrix)$=y_2$$$g_{y_1}...Read more

I was reading Koller's book on Probabilistic Graphical Models and was wondering what the decomposition, weak union and contraction properties of conditional probability mean.But before I ask exactly what I am confused about let me introduce some of Koller's notation so that we are all in the same page (anything else is unclear feel free to ask in the comments). Let capital non-bold stand for random variables say $X$ is a r.v. Let little non bold stand for the assignment to a random variable say $(X = x)$. Also, let me define captial bold letter...Read more

Say I have two independent variables $X$ and $Y$ that are exponentially distributed with respective rates $\lambda_X$ and $\lambda_Y$. How do I compute $\mathbb{E}[X\mid \min\{X,Y\}]$?...Read more

Suppose $X_n$ is the fortune of a gambler after $n$ th game. Then the game is called fair (Breiman 1968) if $$E[X_{n+1} \mid X_1, \dots, X_n] = X_n \forall n$$My question is why a fair game is not defined as the following $$E[X_{n+1}] = E[X_n] \forall n$$ i.e. $$E[X_{n+1}- X_n]=0$$. This should be the proper definition as a fair game is where avg. gain is zero. Nothing conditioning should be there....Read more

Imagine we have $X$ and $Y$ defined on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that $\mathbb{E}|X|, \mathbb{E}|Y|, \mathbb{E}|XY| < \infty$. Let $\mathcal{A}$ and $\mathcal{B}$ be be $\sigma$-fields such that $\mathcal{A} \subset \mathcal{B} \subset \mathcal{F}$.Assume that $X$ has the property that $\mathbb{E}(X| \: \mathcal{B}) = \mathbb{E}(X| \: \mathcal{A})$ and $Y$ has the property that $\mathbb{E}(Y| \: \mathcal{B}) = \mathbb{E}(X| \: \mathcal{A})$ Must it be the case that $\mathbb{E}(XY| \: \mathcal{B}) = \ma...Read more

Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a probability space.Let $$ X, Y : \Omega \rightarrow \mathbb{R} $$ be random variables.Furthermore, let$$ f: \mathbb{R}^2 \rightarrow \mathbb{R} $$be a $\mathcal{B}(\mathbb{R}^2)/\mathcal{B}(\mathbb{R})$-measurable function such that, for all $y \in \mathbb{R}$, the random variables $f(X,y)$ and $f(X,Y)$ have finite expectation.Now let $y \in \mathbb{R}$ be arbitrary. Under the above assumptions, the expected value $\mathbb{E}[f(X,y)]$ and a $\mathbb{P}$-unique conditional expectation $\mathbb{E}[f(X...Read more