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2 votes

Why is average of a random subset the same of the initial set?

Here are two approaches. (1) Here is the calculation using just the definition of expectation. Let $X$ be $\{x_1,x_2,\dots,x_n\}$. Set $Y$ is a random subset of size $l$. We assume all subsets are ...
Aig's user avatar
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-1 votes

Player in casino

In the casino, the player is at the center of the action, navigating through a world of chance and excitement. With every spin of the wheel or flip of the card, anticipation fills the air as the ...
Emily Mia's user avatar
0 votes

Expected payout riddle: How to keep signature collectors from forging?

If I'm understanding this correctly, the agent gets back the $x$ they paid if they successfully complete a job, correct? If that is the case, here's my approach to the solution. If the agent forges 0 ...
math kuma's user avatar
0 votes

Probability that a specific card appears before turning up exactly two aces when turning cards from a random deck?

X is counting the number of cards being turned up, including the two aces in question. So the correct expression for $X$ should be: $$X = 2 + X_1 + X_2 +... X_{48}$$
MathematicsBeginner's user avatar
1 vote

Can Chernoff Bound Theorem be applied to functions of independent random variables

I think McDiarmid inequality is what you are looking for. https://en.wikipedia.org/wiki/McDiarmid%27s_inequality
Ibra's user avatar
  • 140
1 vote

Simple question about expectation

If $X$ is discrete and $P(X>0)>0$ then there exists some $x_i>0$ such that $P(X=x_i)>0$. As $P(X\geq 0)=1$, $X$ cannot take on a negative value with a non-zero probability, because the sum ...
Julio Puerta's user avatar
  • 4,561
0 votes

Expected payout riddle: How to keep signature collectors from forging?

If the agent doesn't forge any signatures, s/he makes $200$. We can compare that with forging one signature. In that case, if the checked signature is not forged the agent makes $202$ If the ...
Ross Millikan's user avatar
2 votes
Accepted

Probability: Question about expected value with coins

None of the coins are more likely to be in any order than the other would you agree? So if we have 2 one dollar coins and 1 five dollar coin, we have FOO, OFO, OOF How many times does each coin show ...
Otis's user avatar
  • 84
1 vote
Accepted

Expected value and average value problem

If $$1-\Pr\left\{ 0.64\leq X\leq0.66\right\}$$ $$=1-\Pr\left\{ \dfrac{0.64-\mu}{\sigma/\sqrt{n}}\leq\dfrac{X-\mu}{\sigma/\sqrt{n}}\leq\dfrac{0.70-\mu}{\sigma/\sqrt{n}}\right\},$$ then $$\Pr\left\{ 0....
AOS's user avatar
  • 181
2 votes
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Order statistics vs Integration to calculate expectation

All things reconsidered, I take back all the words that I said and am about to introduce the solution which works fine and satisfies Python simulations I've run. To calculate the conditional ...
Egor Larionov's user avatar
2 votes

Expected Number of Steps To Reach 0 in Circle

$$E_1-E_2/2=1$$ $$-E_1/2+E_2-E_3/2=1$$ $$-E_2/2+E_3-E_4/2=1$$ $$\dots$$ $$-E_{97}/2+E_{98}-E_{99}/2=1$$ $$-E_{98}/2+E_{99}=1$$ Adding the $99$ equations together gives $E_1/2+E_{99}/2=99$. By symmetry ...
Parcly Taxel's user avatar
3 votes

Calculate $E(X^2)$ from $E(X)$?

As others have shown, you can calculate $E(X^2)$ from the original data but this is not what you asked in the title. You cannot calculate $E(X^2)$ from $E(X)$ without knowing the original data. Let'...
badjohn's user avatar
  • 8,474
2 votes
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Calculate $E(X^2)$ from $E(X)$?

The expected value of $X^2$ is defined as $$E(X^2) = \sum_{k=1}^\infty P(X=k)k^2 = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^kk^2$$ In order to calculate this series, notice that it is the same as $g(...
Julio Puerta's user avatar
  • 4,561
1 vote

Calculate $E(X^2)$ from $E(X)$?

Now that you've clarified that you're talking about the geometric distribution for $p=0.5$, the calculation is simple. $$E(X) = \sum_{k=1}^\infty P(X=k)k = \sum_{k=1}^\infty 0.5^kk = 2$$ $$E(X^2) = \...
Dan's user avatar
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3 votes
Accepted

Randomly choose $n$ numbers between $0$ and $1$. Add the smallest number from each repetition until the total is greater than 1. How many repetitions?

I don't think you can find a closed form solution for any $n$, but for small $n$ the closed form solution can be found: Like that answer you referenced, we first need the probability that the sum of ...
Ben's user avatar
  • 674
3 votes
Accepted

$T$ Pokemon trainers catch a Pokemon every day. How many days does it take until two trainers own Pokemons of the same species?

After $d$ days, there are $Td$ Pokemon caught, and $\binom T2d^2$ pairs of Pokemon owned by different trainers. Each pair has a $1/P$ chance of being the same kind of Pokemon. These collisions are ...
Misha Lavrov's user avatar
1 vote

Randomly choose $n$ numbers between $0$ and $1$. Add the smallest number from each repetition until the total is greater than 1. How many repetitions?

Not a full answer... Let $R(x)$ be the expected number of tries to get an accumulated value that exceeds $x$. Then we can write the integral equation $$R(x) = 1 + \int_0^x R(x-u) \,g(u)\, du = 1 + \...
leonbloy's user avatar
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1 vote
Accepted

$Cov(\max(x, 0), \min(x, 0))$ where $x \sim N(0, 1)$

$$E[Max(X,0) + Min(X,0)] = E[X+0]=E[X]=0$$ Let $f(x)$ be the density of the standard normal distribution. Let $u=-x^2/2, \frac{du}{dx}=-x$ \begin{align} E[max(X,0)] &= \int_{-\infty}^\infty \max(x,...
Siong Thye Goh's user avatar
0 votes

How to calculate the difference of two expected values with different distributions?

You can get different things depending on the properties of $f$, for instance if $f$ is Lipschitz. Consider $\pi$ to be an optimal coupling between $p_1,p_2$. Then we have that: \begin{align} |\mathbb{...
Kevin Ro's user avatar
1 vote
Accepted

coin flipping question

Here's an intuitive reason why the answer is $4/3$. Let's frame the problem slightly differently. First, imagine that the players are flipping the coins simultaneously until at least one of them gets ...
Ziv's user avatar
  • 76
1 vote
Accepted

expected value of a two dice rolling game

Denoting with $X$ the payoff you can compute its expected value as:$$\mathbb{E}[X] = \frac{1}{6}\cdot 1\$ + \frac{1}{2} \cdot (-1\$) + \frac{1}{3}\mathbb{E}[X] \rightarrow \frac{2}{3} \mathbb{E}[X] = -...
rcescon's user avatar
  • 208
2 votes

expected value of a two dice rolling game

Let $E$ denoted the expected payoff per roll. Considering only the rolls that either give us a dollar or we lose a dollar, there are $6+18=24$ possibilities. $\frac{6}{24}=\frac{1}{4}$ of those win us ...
Shean's user avatar
  • 925
4 votes
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Quickly putting something to the power of 100 without a calculator (Shortcuts)

You can use the following limit: $$ \lim_{n\to \infty} \left(1-\frac{1}{n}\right)^{n} =e^{-1} $$ because the equation you wish to approximate is in a similar form: $$ \left[\left(1-\frac{1}{50}\right)^...
acat3's user avatar
  • 11.9k
2 votes

coin flipping question

Let $X$ represent the number of coin flips needed for Jay to observe the first heads, and $Y$ represent the number of coin flips needed for John to observe the first heads. So $X$ and $Y$ are iid ...
heropup's user avatar
  • 137k
0 votes

Inequality regarding expectations

If $X \in L^1(\Omega, \mathcal A, \mathbb P)$, and $\mathcal B \subseteq \mathcal A$ is a sub-$\sigma$-field of $\mathcal A$, then $\left\lvert \mathbb E\left[ X \mid \mathcal B\right] \right\rvert \...
Novice's user avatar
  • 4,112
0 votes

Understanding the "Just one more" paradox on a logarithmic scale

The two "expected values" represent two different things. The 1.15 represents the expected value of the additive strategy, where you bet a fixed amount each time (since in this case your ...
Matt's user avatar
  • 163
1 vote

Randomly choose $n$ numbers between $0$ and $1$. Add the smallest number from each repetition until the total is greater than 1. How many repetitions?

This is a partial answer: The CDF of an uniform random variable on $(0,1)$ is given by $$ \begin{cases} 0 \text{ if } x<0\\ x \text{ if } 0 \le x\le 1\\ 1 \text{ otherwise}\end{cases} $$ So the CDF ...
Marco's user avatar
  • 2,507
1 vote
Accepted

Expectation of a Joint Distribution

If you want the mean $2$D point $(E[X],E[Y])$, then your final expression leads to $\left(\frac{1+1+3+3}{4},\frac{2+4+2+4}{4}\right) =(2,3)$. If you want the mean of the product $E[XY]$, then then ...
Henry's user avatar
  • 157k
2 votes

Inequality regarding expectations

It follows immediately by linearity. Specifically, we know that $$E(a X + b Y | \mathcal{H}) = a E(X| \mathcal{H}) + b E(X|\mathcal{H}).$$ Now, observe that $$|E(X|\mathcal{H})| = |E(X^+|\mathcal{H}) -...
Oscar's user avatar
  • 866
1 vote
Accepted

Proving inequality regarding expected error involving covariance and its estimate.

I will assume that $\hat\Sigma$ is the unbiased sample covariance matrix estimator obtained from an i.i.d. normal sample of size $T$. This means that $\hat\Sigma$ is Wishart-distributed: $$(T-1)\hat\...
Nathan L's user avatar
  • 148
1 vote
Accepted

Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\,y\,\exp\left({-\frac{1}{2}\left(x-y\right)^2-\frac{y^2}{2}}\right)\,\mathrm{d}x\mathrm{d}y$

Do the variable change: $$\begin{cases} u &= x-y \\ v &= y \end{cases}$$ We have $du\wedge dv = dx\wedge dy$, hence $$2\pi \langle XY \rangle = {\int_{\mathbb R\times \mathbb R} (u+v) v \exp(-\...
Just a user's user avatar
  • 15.2k
1 vote

Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\,y\,\exp\left({-\frac{1}{2}\left(x-y\right)^2-\frac{y^2}{2}}\right)\,\mathrm{d}x\mathrm{d}y$

$$ \int_{-\infty}^\infty dyy\exp\left(-\frac{y^2}{2}\right) \int_{-\infty}^\infty dxx\exp\left(-\frac{(x-y)^2}{2}\right)= $$ $$ \int_{-\infty}^\infty dyy\exp\left(-\frac{y^2}{2}\right) \int_{-\infty}^\...
Jon's user avatar
  • 5,492
1 vote

Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\,y\,\exp\left({-\frac{1}{2}\left(x-y\right)^2-\frac{y^2}{2}}\right)\,\mathrm{d}x\mathrm{d}y$

$$\int_{-\infty}^{\infty}x\,\exp\left({-\frac{1}{2}\left(x-y\right)^2}\right)\,\mathrm{d}x$$ $$=\int_{-\infty}^{\infty}(t+y)\,\exp ({-\frac{1}{2} t^{2})}\,\mathrm{d}t$$ $$=0+y\int_{-\infty}^{\infty} \...
geetha290krm's user avatar
  • 37.2k
0 votes

Find expected value of a game

Let $E_i$ be the expected number of matches still to be played from a point where A is on a streak of $i$ wins. Then $$E_0=1+\frac13(E_1+2E_0)$$ $$E_1=1+\frac13(E_2+2E_0)$$ $$E_2=1+\frac13(\phantom{...
Parcly Taxel's user avatar
1 vote
Accepted

Stochastic integration: Computing $\mathbb{E}[\exp(- \lambda \int_0^{t∧T_\epsilon}\frac{ds}{B_s^2}) ]$

The trick to being able to use the Optional Stopping Theorem here is to realize that, in the equation $\alpha (\alpha - 1) = 2\lambda$, we can always choose $\alpha < 0$. This implies $Y_t = (B_{t ...
user6247850's user avatar
  • 13.5k
0 votes

2 Players playing a game to guess a number between 1 and 100 with decreasing payout

Okay I'm considering that you two are choosing at random. Let you be $A$ and $B$ be the other friend. I'm using my basic high school combinatorics knowledge. Please correct me if there is a mistake. ...
Gwen's user avatar
  • 1,241
2 votes
Accepted

How to prove that E[X] < E[Y]?

$$\begin{align} \operatorname{E}[Y - X] &= \operatorname{E}[Y - X \mid Y = X]\Pr[Y = X] + \operatorname{E}[Y - X \mid Y > X]\Pr[Y > X] \\ &= \operatorname{E}[Y - X \mid Y - X > 0]\Pr[...
heropup's user avatar
  • 137k
0 votes

Finding the Supremum of $E(y_t)$?

Pick any $t\in\mathbb{Z}$. Then, it holds $$E[y_t]\le \max\{|a|,|b|\}+\alpha E[y_{t-1}]$$ Supposing, there is an initial value $l\in \mathbb{Z}$, such that $E[y_l]=C<\infty$, you find $$E[y_t]\le \...
user408858's user avatar
  • 2,487
1 vote

The Facebook Birthday Problem(Birthday Problem Variation)

When you have $n$ friends on Facebook, the probability that there is a friend's birthday every day is given by $$P_n=\mathbb P \left (\cap_{i=1}^{365} D'_i \right )=\\1-\mathbb P \left (\cup_{i=1}^{...
Amir's user avatar
  • 5,164
2 votes
Accepted

The Facebook Birthday Problem(Birthday Problem Variation)

If you have $n$ friends, the probability that you have a friend with each birthday is $$ \frac{{n\brace 365}\cdot 365!}{365^n}. $$ The notation ${n\brace k}$ refers to the Stirling numbers of second ...
Mike Earnest's user avatar
  • 76.1k
1 vote
Accepted

An (a.s.) continuous process $(X_t)_{t\geq 0}$ is a Brownian motion if $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale

The main idea is to use the following fact: The conditional characteristic function $\mathbb{E}\left[e^{i\lambda X_s}\vert\mathcal{F}_s\right]$ of $X_s$ is equal to $e^{i\lambda\mu - \frac{1}{2}\...
Wilfred Montoya's user avatar
0 votes
Accepted

Is $(e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0}$ a martingale?

$\mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)}$ does hold for $c\in\mathbb{C}$. The proofs are correct.
Wilfred Montoya's user avatar
1 vote

An (a.s.) continuous process $(X_t)_{t\geq 0}$ is a Brownian motion if $(e^{i\lambda X_t + \frac{1}{2}\lambda^2 t})_{t\geq 0}$ is a local martingale

By using martingale-property and taking derivative at $\lambda=0$, we get martingale property for $X$ $$E[X_{t}|\mathcal{F}_{s}]=X_{s},$$ where we can take derivative due to dominated convergence ...
Thomas Kojar's user avatar
  • 3,621
0 votes
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Expected value of the Brownian motion to the power of $n$

Let $s<t$. From $B_t-B_s$ being independent of $\mathcal{F}_s$ and $B_t-B_s \sim \mathcal{N}\left(0, t-s\right)$ follows: \begin{align*} \mathbb{E}\left[\left(B_t - B_s\right)^n\vert \mathcal{F}_s\...
Wilfred Montoya's user avatar
0 votes

Validity of integral formula for $E[|X + Y|] - E[|X - Y|]$

Here is a more concise derivation, a modification of this answer: First verify the identity $$ |x+y|-|x-y|=2[\min(x^+,y^+)+\min(x^-,y^-)-\min(x^+,y^-)-\min(x^-,y^+)].$$ Next, use this result to argue ...
grand_chat's user avatar
1 vote

The "turning-point fraction" of a random sample from a discrete distribution must have expectation less than 2/3?

Requested from comments and building on Misha Lavrov's answer: This uses linearity of expectation on each of the $n-2$ triplets $X_{i-1}, X_i, X_{i+1}$ with $1<i<n$. If $q_3= \sum\limits_x \Pr[...
Henry's user avatar
  • 157k
5 votes
Accepted

The "turning-point fraction" of a random sample from a discrete distribution must have expectation less than 2/3?

We can use the linearity of expectation argument in the discrete case as well, we just have to be a little bit more careful. For each $i$ with $1<i<n$, there are three cases: The values of $X_{...
Misha Lavrov's user avatar
2 votes
Accepted

Expectation value of repeated dice throws

You can adapt your first expression to the "rethrow 6s" case with $$\mathbb{E}[Y] = 1\times P(1) + 2\times P(2) + 3\times P(3) + \cdots + (6+\mathbb{E}[Y]) \times P(6) $$ which will give you ...
Henry's user avatar
  • 157k
0 votes

Expectation value of repeated dice throws

To find the sum: $$ \sum^{\infty}_{n=1}\sum_{i=1}^6\frac{i}{6^n}= 21\sum_{n=1}^{\infty}6^{-n}={21\over5}=4.2 $$ Becuase the sum of$$\sum_{n=1}^\infty6^{-n}=\sum_{n=1}^\infty\left({1\over6}\right)^n = {...
Masd's user avatar
  • 633
0 votes

Expected value of $X_N$ with smallest index $N$ for which $\sum_{i=1}^N X_i$ exceeds $1$ when $X_i$ are uniformly distributed

Another way: \begin{gather*} f_{S_{N-1}|N}(s|n)=ns^{n-2}\int_1^{1+s}dx=ns^{n-1},\quad E[S_{N-1}|N]=\frac{n}{n+1}\\ f_{S_N|N}(s|n)=\int_{s-1}^1nx^{n-2}dx=\frac{n(1-(s-1)^{n-1})}{n-1},\quad E[S_N|N]=\...
Speltzu's user avatar
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