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36

Hint: The probability that an equal number of tails and heads appear is $\large{{2k \choose k} \frac{1}{2^{2k}}}$ The two remaining outcomes (that there are more heads than tails or more tails than heads) are equally likely.


13

Hint: Fair coin $\implies$ Probability of tails occurring more $=$ probability of heads occurring more $= p$, say. Probability of exactly equal number of heads and tails $=1-2p$. Can you find this one?


13

is it possible that the event has in fact zero probability? Yes. For a situation where this always happens, assume that one observes a random number $x$ drawn from the uniform distribution on $(0,1)$. Then the probability to observe $x$ is zero. (Proof: For every interval $I\subseteq(0,1)$ the probability to observe a number in $I$ is the length of ...


12

I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable. EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end). Introduction & proof outline 0 Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable ...


11

Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof. Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and denote by $\lambda$ the ...


10

The proof of this result has nothing to do with probability theory in general or with Borel-Cantelli lemmas in particular, it is only plain-old-deterministic real analysis: Let $(x_n)$ denote a real valued sequence such that $0\leqslant x_n\lt1$ for every $n$, then $\prod\limits_n(1-x_n)=0$ if and only if $\sum\limits_nx_n$ diverges. Proof: If ...


10

Let me get you started: if $(X,Y)$ is i.i.d., $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$ hence $E(u(Y-X))=E(u(X-Y))$ for every $u$ such that the expectations exist. For $u(t)=t^3$ (or any odd function), this yields $E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3)$ hence... Thus, the answer is "Yes, provided $E(|X|^3)$ is finite" (otherwise ...


10

Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn. The probability that the last ball is blue and that the last green comes after the last red is $\dfrac{30}{10+20+30}\times \dfrac{20}{10+20} ...


8

Note that the integral can be written as $$ \int_0^\infty \int_x^\infty f(t)\,\mathrm dt\,\mathrm dx=\int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dt\,\mathrm dx $$ and using Fubini we get $$ \int_0^\infty \int_0^\infty f(t)\mathbf{1}_{t\geq x}\,\mathrm dx\,\mathrm dt=\int_0^\infty \int_0^t f(t)\,\mathrm dx\,\mathrm dt. $$ Here ...


8

There is a difference between mutually exclusive events and independent events (indeed, a very strict difference in the sense that a couple of non-trivial events cannot be both mutually exclusive and independent). Events $A$ and $B$ are mutually exclusive if it is impossible for both of them to occur simultaneously. To borrow from your example, it is ...


8

Then can you help me to show that $\alpha$ is in fact stopping time? Well, not very easily I am afraid, since $\alpha$ is not always a stopping time for the filtration $(\mathcal F_t)_{t\geqslant0}$ defined by $\mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t)$. For a counterexample, consider some Bernoulli random variable $U$ such that $P(U=1)=\frac12$ ...


8

Wouldn't this be needlessly complicating things? Stirling formula tells you that $$\Gamma(x+1/2)\sim(2\pi)^{1/2}(x+1/2)^{x}e^{-x-1/2}$$ and that $$ x^{1/2}\,\Gamma(x)\sim(2\pi)^{1/2}x^{x}e^{-x} $$ hence $$ ...


8

You may assume the first point $A$ at $(1,0)$ and the second point $B=(\cos\phi,\sin\phi)$ being uniformly distributed on the circle. The probability measure is then given by ${1\over2\pi}{\rm d}\phi$. The distance $D:=|AB|$ computes to $2\left|\sin{\phi\over2}\right|$, and we obtain $${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi ...


8

The problem is the claim that all $x^2$ values between $0$ and $4$ are equally likely. That's just not true, except in the sense that each one of them has probability zero. But in any more reasonable sense, it fails. In particular, if all were equally likely, then the probability that the square was between 0 and 2 would be the same as the probability that ...


8

Write $A \setminus B$ as $A \cap B^{c}$.


8

For a Markov process $(X_t)_{t \geq 0}$ we define the generator $A$ by $$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} = \lim_{t \downarrow 0} \frac{P_tf(x)-f(x)}{t}$$ whenever the limit exists in $(C_{\infty},\|\cdot\|_{\infty})$. Here $P_tf(x) := \mathbb{E}^xf(X_t)$ denotes the semigroup of $(X_t)_{t \geq 0}$. By Taylor's formula ...


7

Suppose that I have a set $S$ of white balls. The calculation $$\binom{n}k\binom{n-k}\ell$$ counts the ways to perform the following operation: pick $k$ of the balls in $S$ and paint them red, then pick $\ell$ of the remaining white balls and color them blue. I could get the same results by first picking any $k+\ell$ balls, then choosing $k$ of those to ...


7

Let $S_{100} = \sum_{k=1}^{100}X_k$ denote the winnings after 100 bets. Then $$E(S_{100})= 100E(X)= 25$$ and $$var(S_{100})= \sum_{k=1}^{100}var(X)= 100(27/16)\approx 168.75.$$ As a sum of $100$ independent random variables, $S_{100}$ is approximately normally distributed with mean $25$ and standard deviation $\sqrt{168.75} \approx 12.99$. The ...


7

For concreteness, assume that we have a biased coin that has unknown probability of landing "heads" and we want to give a $95\%$ confidence interval for $p$. Rrepeat the experiment of tossing the coin $n$ times. Let $\bar{X}$ be the sample proportion of heads. We will use $\bar{X}$ to estimate $p$. Note that $\bar{X}$ has mean $p$ and variance ...


7

With such a large number of trials and with $p = 0.5$, a normal approximation to the binomial distribution would also work. if $X \sim \mathrm{Binomial}(n = 10^6, p = 0.5)$, then $$\Pr[X = n/2] = \Pr\left[\frac{X - np}{\sqrt{np(1-p)}} = \frac{n/2 - np}{\sqrt{np(1-p)}}\right] \approx \Pr[-1/\sqrt{n} \le Z \le 1/\sqrt{n}]$$ using continuity correction, where ...


7

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$. How about $T(A)$? In fact $$ T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\} $$ and is not Borel. added Mar 10 Why is $T(A)$ not Borel? ...


7

Your final expression: $$a_k = \frac{1}{5} \cdot\left( \frac{25}{36} \right)^k $$ This correctly gives the probability that your first 6 will occur on the $2k$th roll. So, $a_1$ is the probability that your first 6 is on roll 2; $a_2$ is the probability that your first 6 is on roll 4; etc. In order to find the probability $p$ that your first 6 is on any ...


7

The measure $P_X$ (not $P_x$) is the image measure of $P$ by $X$, defined by the identity $P_X(B)=P(X^{-1}(B))$ for every $B$ in $\mathcal B(\mathbb R)$ (and $X$ being a random variable is exactly the hypothesis one needs to be sure that $P(X^{-1}(B))$ exists). The fact that, for every measurable function $u$ suitably integrable, $$ \int_\Omega u(X)\,\mathrm ...


7

The probability is $0$, assuming that each point is chosen with uniform distribution on the circle. For a right-angled triangle you need two of the points to be the two endpoints of a diameter. The probability for that is obviously $0$, since there isn't any range of possibilities, and the distribution is continuous. For anyone who is not satisfied so ...


6

First, there are things that are much easier given the abstract formultion of measure theory. For example, let $X,Y$ be independent random variables and let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. Are $f\circ X$ and $f\circ Y$ independent random variables. The answer is utterly trivial in the measure theoretic formulation of probability, but ...


6

All you need is the monotone convergence theorem. Note that if you write $\mathbb E(\cdot)$ as $\int (\cdot) d\mathbb P $ and also write $\sum_{i=1}^\infty(\cdot)$ as $\lim_{n\to\infty}\sum_{i=1}^n(\cdot)$, you can rewrite your desired statement as $$\int \lim_{n\to\infty} \sum_{i=1}^n X_i \ d\mathbb P = \lim_{n\to\infty} \int \sum_{i=1}^n X_i \ d\mathbb ...


6

But actually $X_{n} - X$ are also independent. This is because $X$ is a.s. constant by a simple application of Kolmogorov's 0-1 Law, but this is sort of overkill. So let us give a more direct proof. We first refer to the following lemma: Lemma. $X_{1}, \cdots, X_{n}$ are independent if and only if $$ \Bbb{E}[ f_{1}(X_{1})\cdots f_{n}(X_{n}) ] = \Bbb{E} ...


6

First remark : the events "$S_n^X$ hits $A$ before $B$" are negligible. Let's be rigorous. Let $\tau_A := \inf \{n \geq 0: S_n \geq A\}$, and let $\tau_B := \inf \{n \geq 0: S_n \leq B\}$, both with values in $\mathbb{N} \cup \{+\infty\}$. Then you wan to prove that: $$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) > \mathbb{P}_{(S_n^Y)} (\tau_A < ...


6

If $\varphi(y)=E[X\mid Y=y\,]$ is the non-random function of $y$, then $E[X\mid Y\,]$ is defined to be the random variable $\varphi(Y)$.


6

Integrate the pointwise identity $$\mathbf 1_{a < x \leq b,c < y \leq d}=(\mathbf 1_{x \leq b}-\mathbf 1_{x \leq a})\cdot(\mathbf 1_{y \leq d}-\mathbf 1_{y \leq c})=\mathbf 1_{x \leq b,y \leq d}-\mathbf 1_{x \leq b,y \leq c}-\mathbf 1_{x \leq a,y \leq d}+\mathbf 1_{x\leq a,y \leq c}$$ with respect to $P_{(X,Y)}(\mathrm dx\mathrm dy)$.



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