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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.


1

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


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)) + ...


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.


0

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 ...


0

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.


0

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 $|t|\leqslant1$, $$ ...


1

The best solution would be to derive the exact form of the pdf of $U = (X+Y)(X+Z)$. But if the question is more general, or if closed-form solutions are not available, then the way I would proceed would be to: Step 1: Define the joint pdf Define the joint pdf of $(X,Y,Z)$, say $f(x,y,z)$ ... which is simply the product of the 3 standard Normal pdfs: ...


0

Not sure if it helps and it's a bit late but for the case of the $N\times N$ grid in the limit of $N\rightarrow \infty$ the following exact result for the leaf-probability has been derived here based on an equivalence to the Abelian sandpile model: $$\frac{8}{\pi^2}\left( 1 - \frac{2}{\pi}\right) \approx 0.29454$$


2

Let us assume that the random variables $Y_j$ are independent and that one goal of the OP is to prove that, for every nonnegative $s$, $P(X=s)\geqslant P(X=-s)$. Then an elementary counterexample is $r=2$, $(l_1,l_2)=(1,2)$ and $s=1$, then $$ P(X=1)=P(Y_1=-1)P(Y_2=+1)=(1-p_1)p_2, $$ and $$ P(X=-1)=P(Y_1=+1)P(Y_2=-1)=p_1(1-p_2), $$ hence $P(X=1)\lt P(X=-1)$ ...


2

The mistake is that, even if $(S_n)$ and $N$ are independent, $S_N$ and $N$ are not. But you can write $$\displaystyle\Pr(S_N\in A|N=n)= \frac{\Pr(S_N\in A\cap N=n)}{\Pr(N=n)}= \frac{\Pr(S_n\in A\cap N=n)}{\Pr(N=n)}= \frac{\Pr(S_n\in A)\Pr(N=n)}{\Pr(N=n)}= \Pr(S_n\in A)$$because on $\{N=n\}, S_N = S_n$ .


0

Solved! For every $\epsilon > 0$ $\sum_{n \geq1}\mathbb{P}(Y_n < -\epsilon) < \infty$ then $\mathbb{P}(\lim \sup\{ Y_n< -\epsilon \})= 0$ for Borel Cantelli 1. This is equivalent to $\mathbb{P}(\lim \inf\{ Y_n \geq -\epsilon \})= 1$. In addition $\sum_{n \geq1}\mathbb{P}(Y_n \leq \epsilon) = \infty$ then $\mathbb{P}(\lim \sup\{ Y_n \leq ...


1

Equation 1 follows directly from the definition of expected value. If your random variable $\xi$ has density $p$, the expected value of a function $f(\xi)$ is defined to be $$ \int f(x)p(x)\,dx. $$ Thus $$ M(g(\xi)/p(\xi)) = \int \frac{g(x)}{p(x)}\,p(x)\,dx = \int p(x)\,dx. $$ The expected value of $g(\xi)^2/p(\xi)^2$ is $$ \int ...


0

You may verify the result by calculating it numerically: $E(XY)=\sum_{i,j=1,2,3,4} X_iY_j P(X_i)P(X_j)$ So you sum up all 16 combinations of $X\cdot Y$ weighted by their probability product. By this I came to $E(XY)=6$. As $E(X)=3, E(Y)=2$, we have indeed: $Cov(X,Y)=E(XY)-E(X)E(Y)=6-3\cdot 2=0$


1

$X, Y$ are independent $\implies E[XY] \equiv E[X] \cdot E[Y] \iff E[XY]-E[X]E[Y] \equiv 0 \implies \boxed{\rm{Cov}(X,Y) \equiv0 \ }$


1

Observe that $$ \sum_{k=0}^\infty \mathbf{1}_{(1/2^{k+1},1/2^k)}=\mathbf{1}_{[0,1]},\quad\lambda\, \text{- a.s.} $$ since they agree except in countably many points. Since $X_n$ is non-negative, $|X_n|=X_n$, and $$ \int_{[0,1]}\sup_n |X_n|\,\mathrm d\lambda=\int_{[0,1]} \sup_n X_n\mathbf{1}_{[0,1]}\,\mathrm d\lambda=\sum_{k=0}^\infty\int_{[0,1]}\sup_n ...


2

By definition, $P(X\lt A(T)\mid T)=B(T)$, where, for every bounded measurable function $C$, $$ E(C(T)\,\mathbf 1_{X\lt A(T)})=E(C(T)B(T)). $$ With no further indication on the joint distribution of $(X,T)$, it is difficult to go further. If $(X,T)$ was independent, then $$ E(C(T)\,\mathbf 1_{X\lt A(T)})=E(C(T)\Phi(A(T))), $$ where $\Phi$ is the standard ...


1

This shows why it is important to keep track of when the pdf is zero. Using indicator functions, the pdf of $(X,Y)$ can be written as $$ f(x,y)=2\mathbf{1}_{0\leq x\leq 1}\mathbf{1}_{0\leq y\leq 1-x} $$ and hence $$ \begin{align} P(X<\tfrac12,Y>\tfrac12)&=\iint_{\mathbb{R}^2}f(x,y)\mathbf{1}_{x<1/2}\mathbf{1}_{y>1/2}\,\mathrm dy\,\mathrm dx\\ ...


1

Cauchy Schwarz is fine. For the second equation $\def\Var{\mathop{\rm Var}}\def\Cov{\mathop{\rm Cov}}$(which will hold only if not only the $X_i$, but the pairs $(X_i,X_j)$, $i \ne j$ are identically distributed) recall that \begin{align*} \Var\sum_i X_i &= \Cov\left(\sum_i X_i, \sum_j X_j\right)\\ &= \sum_{i,j} \Cov(X_i, X_j)\\ ...


1

I know equation number 4 is the cauchy schwartz inequality where they are defining their inner product of over this functional vector space as $$\langle f(x),g(x)\rangle=\int^{b}_{a} \left|f(x)g(x)\right|dx$$ and the cauchy schwartz inequality is just $$\left(\langle f(x),g(x)\rangle\right)^{2}\leq \langle f(x),f(x)\rangle\langle g(x),g(x)\rangle$$ As for ...


2

Write $$H_A=\sum_{n=0}^\infty \prod_{k=0}^n 1_{(X_k\notin A)}=1_{(X_0\notin A)}\left[1+\sum_{n=1}^\infty \prod_{k=1}^n1_{(X_k\notin A)}\right]=1_{(X_0\notin A)}[1+H_A^\prime],$$ where $H_A=f(X_0,X_1,X_2,\dots)$ and $H_A^\prime=f(X_1,X_2,X_3,\dots)$. Therefore \begin{eqnarray*} \mathbb{E}(H_A\mid X_1=j,X_0=i) &=&1+\mathbb{E}\left(H_A^\prime \mid ...



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