# Tag Info

84

As has been noted, the usual definition of the Lebesgue integral has little to do with probability or random variables (though the notions of measure theory and the integral can then be applied to the setting of probability, where under suitable interpretations it will turn out that the (Lebesgue) integral of (a certain) functions corresponds to the ...

81

This is a great question to answer graphically. First note that the two can't arrive after 12:45, since they have to spend at least 15 minutes in the bar. Second, note that they meet if their arrival times differ by less than 15 minutes. If we plot the arrival time of person 1 on the x axis, and person 2 on the y axis, then they meet if the point ...

66

I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to be rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts ...

64

$$\text{chance they meet} = \frac{\text{green area}}{\text{green area + blue area}} = \frac{45^2 - 30^2}{45^2}$$

61

This problem, known as the Monty Hall problem, is famous for being so bizarre and counter-intuitive. It is in fact best to switch doors, and this is not hard to prove either. In my opinion, the reason it seems so bizarre the first time one (including me) encounters it is that humans are simply bad at thinking about probability. What follows is essentially ...

58

Maybe this simple example will help. I use it when I teach conditional expectation. (1) The first step is to think of ${\mathbb E}(X)$ in a new way: as the best estimate for the value of a random variable $X$ in the absence of any information. To minimize the squared error $${\mathbb E}[(X-e)^2]={\mathbb E}[X^2-2eX+e^2]={\mathbb E}(X^2)-2e{\mathbb ... 48 We see that in probability, we represent the event A as a set of elements in our sample space, and \neg A as the complement of A in our sample space. Thus, in probabilistic terms,$$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$by the definition of the complement. And by the Kolmogorov Axioms, we see that$$P(\emptyset) = ...

46

(This answer takes as its starting point the OP's question in the comments, "Let me understand mass before going to density. Why do we call a point in the discrete distribution as mass? Why can't we just call it a point?") We could certainly call it a point. The utility of the term "probability mass function," though, is that it tells us something about ...

43

For non-negative integers $a, b$ and $t \in [0, 1]$, the expression $t^a (1 - t)^b$ describes the probability of randomly selecting $a+b$ real numbers in $[0, 1]$ such that the first $a$ are in $[0, t]$ and the last $b$ are in $[t, 1]$. The integral $\int_0^{1} t^a (1 - t)^b dt$ then describes the probability of randomly selecting $a+b+1$ real numbers such ...

39

To understand why your odds increase by changing door let us take an extreme example first. Say there are $10000$ doors. Behind one of them is a car and behind the rest are donkeys. Now, the odds of choosing a car is $1\over10000$ and the odds of choosing a donkey are $9999\over10000$. Say you pick a random door which we call X for now. According to the ...

38

The geometric approach works. Let’s compute the volume of the $2n$ dimensional ball, $D^{2n}$, in two ways. One way is extremely clever but has been known for centuries and provides interesting insights: it’s based on Liouville’s trick. Specifically, we will compute two integrals in polar coordinates, one of which is the volume of the ball and the other ...

37

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.

35

The simplest and surest way to compute the distribution density or probability of a random variable is often to compute the means of functions of this random variable. In the case at hand, one wants to write $\mathrm E(g(W))$ as $$\color{blue}{\mathrm E(g(W))=\int g(w)f(w)\mathrm{d}w},$$ for every bounded measurable function $g$, then one can be sure that ...

30

See Durrett, Probability: Theory and Examples (link goes to online copy of the fourth edition). On p. 164 Durrett gives a proof that simple random walk is recurrent in two dimensions. First find the probability that simple random walk in one dimension is at $0$ after $2n$ steps; this is clearly $\rho_1(2n) = \binom{2n}{n}/2^{2n}$, since $\binom{2n}{n}$ is ...

30

For probability theory as probability theory (rather than normed measure theory ala Kolmogorov) I'm quite partial to Jaynes's Probability Theory: The Logic of Science. It's fantastic at building intuition behind the rules and operations. That said, this has the downside of creating fanatics who think they know all there is to know about probability theory. ...

29

Let's say the $\sigma$-algebra on $X$ is generated by the sets $A_i \subseteq X$. For each subset $I$ of the natural numbers, consider the set $B_I = \bigcap_{i \in I} A_i \cap \bigcap_{i \notin I} (X \setminus A_i)$. For distinct sets $I$ and $J$, the corresponding sets $B_I$ and $B_J$ are disjoint. Now take cases: either only finitely many of the $B_I$ ...

28

This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$: \begin{align*} P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\ &= \sum_{i=0}^k P(Y = k-i , X =i)\\ &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\ &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\ ...

27

I think this is a slightly more intuitive way of looking at the question: Suppose $f (x)=k$. Once we have chosen $k$, there are $200$ possible values for $g (x)$, one of which is $k$, hence we get an answer of $\frac{1}{200}$.

25

A Markov chain is a discrete random process with the property that the next state depends only on the current state (wikipedia) So $P(X_n | X_1, X_2, \ldots X_n-1) = P(X_n | X_{n-1})$. An example could be when you are modelling the weather. You then can take the assumption that the weather of today can be predicted by only using the knowledge of ...

23

Gambling is a good starting-point for probability. We can treat $\sigma$-field as a structure of events as we need to define the addition and multiplication for numbers. The completeness of the real numbers is suitable for our calculations, and $\sigma$-field plays the same role. I hope the following gambling example helps you to understand the filtration ...

23

There's a nice approach in Coverage by Randomly Deployed Wireless Sensor Networks by Wan and Yi that allows us to get an expansion in $\mu:=R/r$ for the expected number of raindrops required. The basic idea is to focus on the intersections among the circles. Since any uncovered area must be bounded by arcs bounded by intersections that aren't covered by ...

23

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 ...

22

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: $$f( (x_1,y_1), ..., (x_n,y_n) ) = \begin{cases}\pi^{-n}& (x_1^2 + y_1^2 < 1) & \text{ & } \cdots \text{ & }&(x_n^2 + y_n^2 < 1) \\ 0 ... 21 I think a good way to answer question 2 is as follows. I am performing an experiment, whose outcome can be described by an element \omega of some set \Omega. I am not going to tell you the outcome, but I will allow you to ask certain questions yes/no questions about it. (This is like "20 questions", but infinite sequences of questions will be allowed, ... 21 Consider$$f(x)=\begin{cases}0&\text{if }x<0\\1&\text{if }x\ge 0\end{cases}$$Any continuous g with ||f-g||_\infty<\frac 13 must have g(x)<f(x)+\frac13=\frac13 for all x<0. By countinuity, g(0)\le \frac13, contradicting g(0)>f(0)-\frac13=\frac 23. Even if we only require |f(x)-g(x)|<\frac13 for almost all x, the ... 21 I absolutely love this result, I literally cannot stop from getting goosebumps and smiling whenever I think about it. It is a proof from probability theory! I learned it in David William's Probability with Martingales, of which it is part of exercise E4.2. Fix s>1 and recall that \zeta(s) = \sum_{n \in \mathbb{N}} n^{-s}, so we aim to show that ... 20 From the way you write the information, it seems that you assume you have only one parameter to estimate (\theta) and you consider one random variable (the observation X from the sample). This makes the argument much simpler so I will carry it in this way. You use the information when you want to conduct inference by maximizing the log likelihood. That ... 20 For a Markov process (X_t)_{t \geq 0} we define the generator A by$$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}$$whenever the limit exists in (C_{\infty},\|\cdot\|_{\infty}). Here P_tf(x) := \mathbb{E}^xf(X_t) denotes the semigroup of (X_t)_{t \geq 0}. By Taylor's formula ... 20 Consider all of the 6\times 5 ways to pick two pieces of fruit. That's 30:$$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ...

19

The problem arises in the countable union; your argument is correct as far as it goes, but from the fact that $\cup_{i=1}^n x_i\in \cup_{i=1}^{\infty}F_i$ for each $n$ you cannot conclude that $\cup_{i=1}^{\infty} x_i$ lies in $\cup_{i=1}^{\infty} F_i$: the full union must be in one of the $F_j$ in order to be in $\cup_{i=1}^{\infty}F_i$. For an explicit ...

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