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2

A simple example is $X$ uniformly distributed on the usual Cantor set, in other words, $$X=\sum_{n\geqslant1}\frac{Y_n}{3^n},$$ for some i.i.d. sequence $(Y_n)$ with uniform distribution on $\{0,2\}$. Other examples are based on binary expansions, say, $$X=\sum_{n\geqslant1}\frac{Z_n}{2^n},$$ for some i.i.d. sequence $(Z_n)$ with any nondegenerate ...


0

The iff is not true: $$f(x)=\begin{cases} 2 \text{ if } 0\le x \le \frac {1}{4}\\ \frac {2}{3} \text{ if } \frac {1}{4}< x \le 1\\ 0 \text{ otherwise}\end{cases}$$ $$ f\not\in C^0$$ $$F_X(x)=\int_{-\infty}^x f(x) dx$$ $X$ is continuous r.v.


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As explained in the comments, this is really the iteration of a transformation of distributions, not random variables, and the technique used to find the fixed points in the other question can be adapted to solve the asymptotics. To be brief (since this is a rehashing of arguments already explained), the sequence of generating functions ...


3

I believe what you claim only holds for $\sigma<1$ I am going to truncate the expectation at a large positive $c$, i.e. We will just look at the tail of the the integral. The thing you asked for is finite iff the following integral is finite (by symmetry of $Z$, $\phi$ and $\Phi(\cdot)(1-\Phi(\cdot))$, we do not need to look at the tail at $-\infty$). ...


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Define $\triangle=\left\{ \left(x,x\right)\mid x\in\mathbb{R}^{2}\right\} $ here. Suppose that $f\left(x,y\right)$ serves as PDF here. Then we get the contradiction: $$1=P\left\{ \left(X,X\right)\in\mathbb{R}^{2}\right\} =P\left\{ \left(X,X\right)\in\triangle\right\} =\int_{\triangle}fd\lambda=0$$ The last equality as a consequence of ...


1

Both integral representations are applications of the following: If $Y$ is a random variable with values in $\mathbb{R}^n$, then $$ {\rm E}[u(Y)]:=\int_{\Omega} u(Y)\,\mathrm dP=\int_{\mathbb{R}^n}u(y)\,P_Y(\mathrm dy),\tag{1} $$ for measurable, bounded $u$. Here $P_Y$ is the distribution of $Y$ which is the measure on $\mathbb{R}^n$ given by $$ ...


2

This is known as Vandermonde's identity. A combinatorial interpretation of the identity follows: Suppose you have $n$ boys and $m$ girls. The RHS counts the number of ways to choose $k$ children out of the total $n+m$ children regardless of sex; the LHS counts the number of ways to choose $i$ girls and $k-i$ boys for each $i = 0, 1, 2, \ldots, k$. ...


2

Hint: use Poisson distribution with $\lambda_1 = 4, \lambda_2 = 12$.


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Found it finally in this paper. Let $U_1$ and $U_2$ be independent uniforms on $(0, 1)$. Then the standard Box-Muller algorithm tells us that the random variables: $$ V_1 = \sqrt{-2 \log U_2} \cos(2\pi U_1) $$ $$ V_2 = \sqrt{-2 \log U_2} \sin(2\pi U_1) $$ are independent $N(0, 1)$. Let $W_1$ and $W_2$ be correlated normals: $$ W_1 = V_1 $$ $$ W_2 = \rho ...


0

A random variable $f:O\to\mathbb R$ is an $F$-measurable function, meaning for every $\alpha \in \mathbb{R}$ $$\{\omega \in O: f(\omega) > \alpha\} \in F.$$ Since $F$ is a $\sigma$-algebra, other common definitions are equivalent: $$\{\omega \in O: f(\omega) \geq \alpha\} \in F, \\\{\omega \in O: f(\omega) < \alpha\} \in F,\\\{\omega \in O: ...


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$f^{-1}(x)$ would be a member of the uunderlying set $O$, not a subset of $O$. Thus the question of whether it is a member of $F$ doe not arise. Maybe you meant the set $$ f^{-1}(\{x\}) = \{\omega\in O : f(\omega)=x \}. $$ By the usual definitions, $f$ would not be a random variable on this spacec unless that set is a member of $f$ for every $x\in\mathbb ...


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Hint: Add the domain information about the density functions as unit step functions to the integration and take the integral on the whole real line. This will give you the correct result. E.g.: $f(x)=\frac{1}{x^2}{\bf 1} _{\{1/(a+1),1/a\}}$


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For a geometric proof, see Bennett Eisenberg & Rosemary Sullivan, Why Is the Sum of Independent Normal Random Variables Normal?, Mathematics Magazine, Dec. 2008, 362-366, available at http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/why-is-the-sum-of-independent-normal-random-variables-normal


1

Your attempt breaks down at the second = sign when some mysterious $t$ appears (what is $t$ here?). Instead, one should write $$ (\ast)=P(T_n \leqslant s <T_{n+1})=P(T_n\leqslant s\lt T_n+\tau_{n+1}). $$ Now, $(T_n,\tau_{n+1})$ is independent and one knows the distributions of $T_n$ and $\tau_{n+1}$ hence one can evaluate the RHS. Say the distribution of ...


1

As Peter stated, to calculate the mean, you multiply each outcome (the number that you roll) by the probability of getting that roll, and add those products up. On a regular or fair die, yes, the probabilities of each outcome are the same. But this isn't a fair die; $6$ is twice as probable as each of the other numbers. For a fair die, $$E(X) = ...


3

The mean of a discrete random variable is defined as $$\mathbb{E}(X)=\sum\limits_{x\in X} xp(x)$$ In this case $X=\{1,2,3,4,5,6\}$, so that's where you are getting the multiplication. You can think of the probabilities as weighting the importance of each of these $6$ numbers.


1

Let $ X \sim {\mathcal U}[-\alpha, \alpha] $ and let $ Y $ be a discrete r.v. taking values $ \pm\alpha $ with equal probabilities. Then $ X+Y \sim {\mathcal U}[-2\alpha, 2\alpha] $. I believe that the sum of $ n \geq 3 $ independent r.v.'s distributed on $ [-\alpha, \alpha] $ cannot be uniform on $ [-n\alpha, n\alpha] $, but I have no proof of this.


2

Actually, one is not interested in $P(Y<c/u(X))$ but in $P(Yu(X)\lt c)$, and this is $$P(Yu(X)\lt c)=\int_{t}^{\infty}f_X(x)F_Y(c/u(x))dx +\int_{0}^{t}f_X(x)dx.$$ Thus, the factor $$ \int_{c/u(x)}^{\infty}f_Y(y)dy $$ in the second part of the RHS is not useful. Note that, if $x\lt t$, $u(x)\lt0$ hence $c/u(x)\lt0$ and, since $f_Y(y)=0$ for every $y\lt0$, ...


1

Hint: $X, Y$ are uncorrelated if $E(X,Y)=E(X)E(Y)$. Compute then $E(X Y) = E[(ab-cd)(cp-aq)]$ using the property that, if $s$ $t$ are independent $E(st)=E(s)E(t)$


0

For your simple example with two dependent Bernoulli random variables with parameters $p_1$ and $p_2$ (where we assume without loss of generality that $p_1 \geq p_2$, and also that $p_1+p_2 \leq 1$), $$P\{X+Y = 2\} = P\{X = 1, Y = 1\} \in [0, p_2]$$ where if $P\{X+Y = 2\}$ has the minimum possible value $0$, then it must be that $$P\{X+Y=2\} = 0, \quad ...


1

You are right that $X$ can still be a random variable. Take $Y$ to be uniformly distributed among $\{-2, -1, 1, 2\}$, and let $X = 1_{Y > 0}$. Then $X$ is Bernoulli($1/2$) on $\{0, 1\}$ and $\sigma(X) \subset \sigma(Y)$. What your professor probably means is that once $Y$ is known, $X$ is "determined." In other words, $\mathbb{E}[X | \sigma(Y)] = X.$


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As I understand the problem you do need to consider the case in which he does not through a six three times in a row since this is when he loses money. The easiest way to approach this kind of problem is to draw some kind of tree in which the end nodes are the possible outcomes. I got that the expected value is about -1.685 (-3 * (125/216)) + (1 * (1/6)) + ...


0

$$\frac{1}{6}-\frac{5}{6}+\frac{5}{6}\cdot \frac{1}{6}-\left(\frac{5}{6}\right)^2+\frac{1}{6} \left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^3=-1.68518$$ He loses 1.68518 rupees. This agrees with a simulation.


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It's still possible for the fourth case (he doesn't get a six on any of his rolls) to happen, correct? So to answer your question, yes, you need to consider the case where he doesn't get any sixes. I think if you work through everything, he should end up expecting to lose about 1.69 rupees.


0

For $z$ in the interval $(0,d)$, the integration is over the interval $\frac{z}{d}\le y\le 1$ . The variable $y$ must be positive and $\le 1$, since $f_Y(y)=0$ outside $[0,1]$. The restriction $y\ge \frac{z}{d}$ comes from the fact that if $y\lt \frac{z}{d}$, then $\frac{z}{y}\gt \frac{z}{z/d}=d$. But then $f_X\left(\frac{z}{y}\right)=0$.


2

In the Lebesgue theory, one defines $EX$ as $EX^+ - EX^-$, where $X^+=\max \{ X,0 \}$ and $X^-=\max \{ -X,0 \}$. Thus the condition is really that $EX^+$ and $EX^-$ are both finite, in which case both $EX$ and $E|X|=EX^+ + EX^-$ exist and are finite.


1

We can prove more general fact. Let $(\Omega,F,p)$ be a probability space where are defined iidrv's $X_k(k=1,2, \cdots)$(in our case $(\Omega,F,p)=(R^N,\cal{E},\mu)$) and $\{ X_k : k \in N\}$ are coordinate projections). The $\theta$-shift invariance means the following: For each Borel subset $A \subset \cal{B}(R^N)$ the following equality $p(\{\omega ...


0

If $X_1,..., X_n$ are independent Poisson random variables having parameter $1$, then $Y=X_1+...+X_n$ is also a Poisson r.v. with parameter $n$. We have also; $\mathbf E(Y)=Var(Y)=n$ The central limit theorem then implies that; $Z:=\displaystyle\frac{Y-\mathbf E(Y)}{\sqrt{Var(Y)}}=\frac{Y-n}{\sqrt{n}}$ $\qquad$$(I)$ is approximately normal and ...


1

Some confusion might stem from neglecting the fact that $\theta^{-1}$ is not defined as a function, at least not defined on the image set of $\theta$. Rather, for every function $\theta:E\to F$ (whether $\theta$ is injective or surjective or not) and every $A\subseteq F$, one defines $\theta^{-1}(A)=\{x\in E\mid\theta(x)\in A\}$. Thus, $\theta^{-1}$ is not ...


0

By Linearity of Expectation $E(W) = 3\times E(U) - 2\times E(D) + 4\times E(N) \times E(B)$ Mean of Uniform Distribution $= \frac{(a+b)}{2}$, for $E(U) = \frac{(2+4)}{2} = 4$ $E(D) = \frac{(3+7)}{2} = 5$ Mean of a Normal = Mean $= 3.5$ Mean of a Bernoulli = $p = 0.65$ Putting all together $E(W) = 3\times4 - 2\times 5 + 4\times 3.5\times 0.65 = 11.1$ ...


0

To give this a simple answer: Yes, the approach described in the question works fine.


2

Yes, the joint cumulative distribution function $F_{X,Y}(x,y)$, that is, the probability that $X\le x$ and $Y\le y$, is the product of the individual cdf $F_X(x)$ and $F_Y(y)$. If $X$ and $Y$ have continuous distributions with density functions $f_X(x)$ and $f_Y(y)$, then the joint density function $f_{X,Y}(x,y)$ is the product of the individual density ...


0

I am answering my own question above. The correct answer is that $\mu \neq \mu_1$ and $\sigma \neq \sigma_1$. Thus, the normalizing constants for the maximum and minimum extreme value distributions are entirely different. This is illustrated in the monograph "Extreme Value and Related Models with Applications in Engineering and Science by Enrique Castillo, ...


1

Rahul's comment exploits a peculiar feature of two-dimensional sphere: the spherical area bounded between two parallel planes is proportional to the distance between them. (See here). This is why we can sample points uniformly on unit 2-sphere by choosing cylindrical coordinates: $z$ uniformly from $[-1,1]$ and longitude $\theta$ uniformly from $[0,2\pi]$. ...


1

$$\mathbb{E}X=\frac{1-p}{2}\times\left(-1\right)+\frac{1}{2}\times0+\frac{p}{2}\times1=\dots$$ $$\mathbb{E}X^{2}=\frac{1-p}{2}\times\left(-1\right)^{2}+\frac{1}{2}\times0^{2}+\frac{p}{2}\times1^{2}=\dots$$ $$\text{Var}X=\mathbb{E}X^{2}-\left(\mathbb{E}X\right)^{2}=\dots$$ Help yourself. Your original solution (before your edit) ...


0

E(X) is correct. And the variance is $Var(X)=E(X^2)-\left[ E(X) \right] ^2$.


0

generate any number k select number (k mod 10), enumerate the remaining 9 numbers in sequence generate any number k select number (k mod 9), enumerate the remaining 8 numbers in sequence etc.


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First generate generate a random number 1 to 9; lets say you get 4 Next generate a random number 1 to 8 Lets say you get 5 so the number you use is the 5th one you have not already used so it would be 6. Next generate a random number 1 to 7 Lets say you get 2 so the number you use is the 2nd one you have not already used so it would be 2. Repeat for all ...


0

What you want is to find a permutation of the set $\{1,2,3,4,5,6,7,8,9\}$. There are $9!$ such permutations in general, and picking one at random will produce an ordering of the numbers. The easiest way to do it is to have a way of selecting a random element from a set. That way, you just perform the process: Pick random number $x$ from $S$ Replace $S$ ...


3

The Fisher-Yates Shuffle gives an efficient way of generating a permutation uniformly at random. From an initial list, it performs $O(n)$ swaps. Afterwards, we can just read the entries from the permutation.


2

Page 2 gives the solution to this: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-436j-fundamentals-of-probability-fall-2008/recitations/MIT6_436JF08_rec05.pdf The crux of the problem is covered in the answer by André Nicolas.


3

Hint: In addition to the linearity property you mentioned, use the following facts: $1$) By symmetry we have $E\left(\frac{X_i}{\sum}\right)=E\left(\frac{X_j}{\sum}\right)$. $2$) $E\left(\frac{\sum}{\sum}\right)=E(1)=1$. This, $1$), and linearity forces $E\left(\frac{X_i}{\sum}\right)=\frac{1}{n}$. Existence is not a problem since $0\lt ...


3

Let $X_{i}$ be i.i.d. and $Y_{n} = \sum_{i=1}^{n} X_{i}$. It is well-known that If $\Bbb{E} X$ exists in $[-\infty, \infty]$, then $S_{n}/n \to \Bbb{E} X$ a.s. If $\Bbb{E}|X| = \infty$, then in probability 1, $S_{n}/n$ does not converge in $(-\infty, \infty)$. (The first case is SLLN when $\Bbb{E}|X| < \infty$, and the case $\Bbb{E}X = \pm\infty$ are ...


0

For each $\omega$ you must have $-X(\omega) < \epsilon(\omega) <1-X(\omega)$.


3

The function $t\mapsto\frac1{1+t}$ is convex on $t\geqslant0$ hence $$ E\left(\frac1{1+X^2}\right)\geqslant\frac1{1+E(X^2)}=\frac1{1+\nu}. $$ The lower bound is attained when $P(X=\sqrt\nu)=P(X=-\sqrt\nu)=\frac12$. On the other hand, if $P(X=0)=1-\frac\nu{x^2}$ and $P(X=x)=P(X=-x)=\frac\nu{2x^2}$ for some $|x|\geqslant\sqrt\nu$, then $E(X)=0$ and ...


0

The answer is that every $X_i$ is Cauchy. I am not sure if you knew these fact: If $Y_1, ... Y_n$ have Cauchy distribution, then $\frac{1}{n}(Y_1+...Y_n)$ is also Cauchy. If $Y_i$ is a (standard) Cauchy distribution, so is $X_i=\frac{1}{Y_i}$. Combining these facts, the $X_i$ you want are Cauchy distributions. (though it is a bit unclear to me what ...


1

Notice that, since $n\theta =\sum \theta_i$ $\sum (\theta_i-\theta)^2= \sum (\theta^2 +\theta_i^2 -2\theta\theta_i)=\sum \theta_i^2 + n\theta^2 - 2\theta\sum \theta_i = \sum \theta_i^2 + n\theta^2 - 2n\theta^2 = \sum \theta_i^2 - n\theta^2$ You should certianly remember this trick. It will crop up again and again...


3

The $A_k$ are independent because for any $i \neq j$, the sets of tosses $2^i, \ldots, 2^{i+1}-1$ and $2^j, \ldots, 2^{j+1}-1$ are disjoint. Hence, $A_i$ tells you no information about $A_j$. Let's find upper and lower bounds for $P(A_k)$. Amongst the tosses numbered $2^k, \ldots, 2^{k+1}-1$, there are $2^k$ tosses, and so, there are $2^k-k+1$ blocks of ...


1

The $A_k$'s are clearly independent, because they involve disjoint sets of coin flips. ($A_0$ depends on flip $1$; $A_1$ depends on flips $2$ and $3$; $A_2$ depends on flips $4$ through $7$; and so on.) So the Borel-Cantelli lemma applies. Also, "$A_k$ infinitely often" is a tail event, so Kolmogorov's zero-one law says that it must have probability $0$ ...


2

All the pieces are there, this is only a matter of using them in the right order. Note first that $$ 1-\Re\varphi_X(u)=\Re(1-\varphi_X(u))=\int_\mathbb R(1-\cos(ux))\mu_X(\mathrm dx), $$ hence $$ 1-\Re\varphi_X(u)\geqslant\int_A(1-\cos(ux))\mu_X(\mathrm dx), $$ because $1-\cos$ is nonnegative everywhere. Next, note that, for every $|t|\leqslant1$, $$ ...



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