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35

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.


21

Given: $(X,Y)$ is Uniformly distributed on a disc of unit radius. Then, the joint distribution of $n$ such points, $((X_1,Y_1), ..., (X_n,Y_n))$, each drawn independently, will have joint pdf: $$f( (x_1,y_1), ..., (x_n,y_n) ) = \begin{cases}\pi^{-n}& (x_1^2 + y_1^2 < 1) & \text{ & } \cdots \text{ & }&(x_n^2 + y_n^2 < 1) \\ 0 ...


16

Here are some asymptotics for the first part: Claim For every fixed $m$, the random variable $$ S_n^m = T_{n}^m - \log n - (m-1) \log\log n+\log((m-1)!) $$ converges in distribution to the "extreme value" distribution independent of $m$ given by $$ F_{S_\infty}(s) = \exp\left(-\mathrm e^{-s}\right). $$ Proof For frog $j$ and $t\ge 0$, let ...


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 ...


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?


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

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} ...


9

As the other respondents have already noted, probability is already defined as a ratio. But here is another thing to think about. Normally when you take the ratio of two quantities, the result is measured in the units of the ratio of the units. For example, if you travel 10 meters in 5 seconds, your average velocity is 10/5 meters per second. But ...


9

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 ...


8

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 ...


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

There is a very elegant theorem published by Beardwood, Halton and Hammersley in 1959 that gives an alsmost-sure convergence for the length of the shortest path divided by $\sqrt{N}$. See this poster for a statement in a general setting. Also, the convergence holds for expectations. In 1989, Rhee and Talagrand showed that the length of the shortest path is ...


8

No, because you're not in a situation where you always remember two other wrong answers at the moment you're about to mark your choice. The analysis in the Monty Hall problem that leads to "it is an advantage to switch" depends on the fundamental assumption that you will be shown a non-prize door that you haven't chosen no matter which choice you made ...


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 ...


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

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

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

Let us formulate your problem mathematically. Let $X_n$ be the outcome of each coin toss. $P(X_n = 1) = 1- p$ and $P(X_n = -1) = p$ Let $H_{n}$ be your bet size on $n$-th toss and this depends on outcomes of previous coin tosses $X_1,... X_n$. Let $S_n = \sum_{i=1}^n X_n$ Then your final wealth is $W_n = W_0+\sum_{i=1}^n H_n X_n = W_0 + \sum_{i=1}^n H_n ...


7

Your answer depends, as you guessed on the process $u \mapsto \sigma_u$. You can amplify your approach using the Ito-isometry with the BDG-inequality: \begin{align} \mathbb{E}[|\int_0^t\sigma_udW_u - & \int_0^s\sigma_u \,dW_u|^{2p}] \stackrel{\text{BDG}}{\leq} c(p) \mathbb{E}[(\int_s^t |\sigma_u|^2 \, du)^p] \\ & \leq c(p) \mathbb{E}[(t-s)^{p-1} ...


7

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 ...


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

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

Integrate the LHS and the RHS of the pointwise identity $$ X=\int_0^X\mathrm dx=\int_0^\infty\mathbf 1_{X\geqslant x}\,\mathrm dx. $$ This shows that the desired formula for $E[X]$ holds irrespectively of the hypothesis that $X$ is discrete or continuous or neither discrete nor continuous, as soon as $X\geqslant0$ almost surely, and that two formulas are ...


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

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

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 ...


7

Denote by $\Phi(t) = \mathbb{E}e^{\imath \, t \cdot X}$ the characteristic function of $X$. We have $$X = \frac{1}{2} \big((X+Y)+(X-Y) \big).$$ Thus, $$\begin{align*} \Phi(t) &= \mathbb{E}e^{\imath \, \frac{t}{2} (X+Y)} \cdot \mathbb{E}e^{\imath \, \frac{t}{2} (X-Y)}= \left( \mathbb{E}e^{\imath \, \frac{t}{2} X} \right)^2 \cdot \mathbb{E}e^{\imath \, ...


7

Watch the phrase "almost surely". Here is a written-out statement of the problem: Suppose that for each $n$, there exists an event $A_n$ with $P(A_n) = 1$ such that for every $\omega \in A_n$, we have $|X_n(\omega)| \le Y(\omega)$. Show that there exists an event $A$ with $P(A) = 1$ such that for every $\omega \in A$, we have $\sup_n |X_n(\omega)| \le ...


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? ...


6

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 ...



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