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6

For $i=1$ to $50$, define random variable $X_i$ by $X_i=1$ if box $i$ has no balls, and $ X_i=0$ otherwise. Then $X=X_1+\cdots +X_{50}$, so we want $E(X_1+\cdots+X_{50}$). By the linearity of expectation this is $E(X_1)+\cdots+E(X_{50})$. But $\Pr(X_i=1)$ is the probability all the balls miss box $i$. This is $\left(\frac{49}{50}\right)^{100}$, and is ...


3

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 ...


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 $-\infty$). ...


3

Let $Y$ be uniform random variable on $(0,1)$ and conditionally on $Y$, $X$ is a centered normal random variable with variance equal to $Y$. Then $E(X|Y) = E(X) = 0$, but $X$ is not independent of $Y$ since $E(X^2 | Y) = Y$ $E(X^2) = E(Y) = \frac{1}{2}$ implies $X \in L^1$


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.


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 $y\lt0$, ...


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 $\lambda_1 = 4, \lambda_2 = 12$.


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 ...


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

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)) + ...


1

Your result is correct. It can be done more efficiently by: $$P\left\{ z<X<Y\right\} =\int_{z}^{\infty}f_{X}\left(x\right)P\left\{ z<X<Y\mid X=x\right\} dx=\lambda\int_{z}^{\infty}e^{-\lambda x}e^{-\mu x}dx=\frac{\lambda}{\lambda+\mu}e^{-\left(\lambda+\mu\right)z}$$


1

Prob$(X=i)$=Prob(All 100 balls go into $(50-i)$ boxes)=${50\choose{50-i}}*\frac{(50-i)^{50+i}}{50^{100}}$ The denominator is the total number of ways 50balls can be put in 100 bxes, each ball can be put on any of the 50 boxes, i.e in 50 ways, so 100 balls can be put in $50^{100}$ ways. The numerator is the product of number of ways to choose the (50-i) ...


1

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

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 $$ ...


1

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

Consider independent Poisson processes $N_t$, $M_t$ with rates $\lambda$, $\mu$, such that $X$ is the time until the first occurrence of $N_t$ and $Y$ is the time until the first occurrence of $M_t$. Then $N_t + M_t$ is a Poisson process with rate $\lambda + \mu$, and $Z$ is the time until its first occurrence. One way to realize this is to start with a ...


1

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 ...


1

$E[Y] = E[E[Y|X]]$ The conditional expectation is $pX$. Then, $E[pX] = pE[X] = \mu p$


1

Formally, whenever you have a parametrized family of distributions $D(\theta)$ for the parameter set $\theta\in \Theta$ and some random variable $X$ which takes values in $\Theta$ and has the distribution $q$, if you define a new random variable as $Y \sim D(X)$, then the distribution of $Y$ is given by $$ \mathsf P(Y\in A) = \int_\Theta ...


1

Define $\xi_i$ to be a family of independent identically distributed Bernoulli variables with success probability $p$. Then $E[\xi_i]=p$ for all $i$. Note that $$Y=\sum_{i=1}^X\xi_i $$ Wald's Identity states that if $N$ is positive integer-valued random variable with finite expectation and $\eta_i$ are independent identically distributed random ...


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) = ...


1

If $y=0$, then $Y_y=0$ or $Y_y=1$, depending on the convention. If $y\gt0$, then: $P(Y_y\gt0)=pP_1(T_y\lt T_0)$ $P(Y_y\gt n+1\mid Y_y\gt n)=P_y(T_y\lt T_0)=pP_{y+1}(T_y\lt\infty)+(1-p)P_{y-1}(T_y\lt T_0)$ Similar recursions hold when $y\lt0$. If one is able to compute the various probabilities $P_a(T_b\lt T_c)$ involved (and one should be), these ...


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 ...


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

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 ...


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, $\mathbb{E}[X | \sigma(Y)] = X.$


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 $E(st)=E(s)E(t)$


1

Hint Find expressions for $$P\left\{ \left|X\right|\leq x\wedge\frac{X}{\left|X\right|}=1\right\} =P\left\{ 0<X\leq x\right\} $$ and $$P\left\{ \left|X\right|\leq x\right\} P\left\{ \frac{X}{\left|X\right|}=1\right\} $$ and compare. If they are the same then you are ready. Use the symmetry of CDF $\Phi(x)$ of $X$


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 $[0,2\pi]$. ...



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