New answers tagged

0

If $X\sim\mathcal{N}(6,4)$, then using the properties of the normal distribution, we have $\frac{X-6}{\sqrt{4}}\sim\mathcal{N}(0,1)$. Thus $P(X>a)=0.4\iff P\left(\frac{X-6}{2}>\frac{a-6}{2}\right)=0.4\iff P\left(\frac{X-6}{2}\leq \frac{a-6}{2}\right)=0.6$ As $\frac{X-6}{\sqrt{4}}\sim\mathcal{N}(0,1)$, this is equivalent to ...


2

There are two ways of looking into this problem. For this we need to define what uniform distribution means. I will alter and observe the interval $[0,1]$ and adopt $x = [x]+\{x\}$ for clarity. If we take that uniform distribution is corresponding to uniform distribution of random numbers then $[\pi^k]$ does not correspond to uniform random distribution. ...


1

Maybe this would get better answers at stats.stackexchange.com than here? Maybe this isn't extremely "real-world", but it's a place where the Cauchy distribution appears where people were not thinking about Cauchy distributions: A Brownian motion in the $(x,y)$-plane starts at $(0,1)$. The $x$-coordinate of the point at which it first reaches the $x$-axis ...


1

We have that $$\begin{align} \operatorname E\mathrm{sgn}(X-Y) & =-1\cdot\Pr(X<Y)+0\cdot\Pr(X<Y)+1\cdot\Pr(X>Y) \\ & =\Pr(X>Y)-\Pr(X<Y) \end{align}$$ and $$\begin{align} \operatorname E\mathrm{sgn}^2(X-Y) & =(-1)^2\cdot\Pr(X<Y)+0^2\cdot\Pr(X<Y)+1^2\cdot\Pr(X>Y) \\ & =\Pr(X>Y)+\Pr(X<Y) \end{align}$$ using the law ...


0

The answer above is called the probability integral transform. Aside from using that, we can directly compute the expectation with integration by parts. First, note that the pdf is the derivative of the cdf. $$ \begin{align*} \mathbb{E}\left[\Phi\left(X\right)\right] &= \int_{-\infty}^{\infty} \Phi\left(x\right)\phi\left(x\right)\, dx\\ &= ...


3

Observe that an element $k$ of a permutation of $S_n$ can participate in zero, one, two etc. up to $k-1$ inversions. Hence we obtain the following generating function of permutations of $S_n$ classified according to inversions: $$G(z) = 1\times (1+z)\times (1+z+z^2)\times\cdots\times (1+z+z^2+\cdots+z^{n-1}).$$ This is $$G(z) = ...


0

Assuming you want to estimate $\sigma$, you need $c=c_n$ to satisfy $\sigma=E(T)$ in order for $T$ to be unbiased. Using rules of expectation and variance, you should get expectation $$ E(T)=E(c\sum|X_i|) = cnE|X_1|, $$ since the $X$'s are identically distributed, and variance $$ V(T)=V(c\sum|X_i|)=c^2V(\sum|X_i|) \stackrel{(1)}= c^2\sum V(|X_i|) ...


0

$$F_X(n)=P_x(X\le n)=\sum_{n_i\le n} 2^{-n_i}= 1-2^{-n}$$ $$1-2^{-n} \geq t\\ 1-t \geq2^{-n} \\ \log_2(1-t) \geq -n \\ n \geq - \log_2(1-t) = \log_2\left( \frac{1}{1-t} \right) $$ $$ F_X^{-1}(t) = \left\lceil \log_2\left( \frac{1}{1-t} \right) \right\rceil $$ where $\lceil x \rceil$ is the least integer greater than or equal to $x$ function.


2

I will illustrate the process of determining what you call the "luck percentile" by a specific case: Sergeant Joe fires ten shots, with success probability of $3/4$ on each shot. He hits nine of the ten targets. Well, if you fire ten shots with probability $3/4$ of success on each, the number of hits will be distributed according to the binomial ...


0

You measure the ratio $r$ $$ r = \frac{\text{hits}}{\text{shots}} \to p = \frac{P}{100} $$ which approaches the probability $p$, expressed as percentage $P$.


3

Hint: Show that $$\mathcal{D} := \{B \in \mathcal{B}(\mathbb{R}); X^{-1}(B) \in \mathcal{F}\}$$ is a Dynkin system. Conclude from the fact that $$\mathcal{G} := \{(a,b]; a<b\}$$ is contained in $\mathcal{D}$ and that $\mathcal{G}$ is a $\cap$-stable generator of $\mathcal{B}(\mathbb{R})$ that $$\mathcal{D} = \sigma(\mathcal{G}) = ...


0

This problem relates to empirical distribution theory. Here, a complete treatment of the subject: EMPIRICAL PROCESSES: Theory and Applications https://www.stat.washington.edu/jaw/RESEARCH/TALKS/Delft/emp-proc-delft-big.pdf


1

This is just the definitions. Distribution $\mu$ is absolute continuous wrt distribution $\nu$ means for any (measurable) set $A$, $\nu(A)=0$ implies $\mu(A)=0$. Does that hold for your examples? No, $A=[3,5]$ has measure zero under $Unif[0,3]$ but 1/2 under $Unif[1,5]$. $\mu$ is singular wrt $\nu$ means for any set $A$, $\mu(A)>0$ implies $\nu(A)=0$. ...


4

\begin{align} \operatorname{var}(Z) & = \operatorname{var}(\operatorname{E}(Z\mid N)) + \operatorname{E}(\operatorname{var}(Z\mid N)) & & (\text{This is the law of total variance.}) \\[10pt] & = \operatorname{var}(N\mu) + \operatorname{E}(N\sigma^2) \\[10pt] & = \mu^2 \operatorname{var}(N) + \sigma^2 \operatorname{E}(N) \\[10pt] & = ...


0

If $T = X_1 + X_2 + \cdots +X_N$, where $X_i$ are iid and $N$ is independent of $X_i$, then a standard result, usually derived via a conditioning argument, is that $E(T) = E(N)E(X)$ and $V(T) = E(N)V(X) + V(N)[E(X)]^2.$ (Search the Internet for proofs under 'Random sum of random variables'; the UNL page seems complete.) A small adjustment will deal with ...


1

The sequence is almost surely non-convergent. Assume the sequence was convergent. By the Hewitt-Savage zero-one law the limiting random variable has to be constant. If the limit isn't zero, we can apply the continuous mapping theorem and we get that $\log(R_n)$ converges almost surely to a constant. This is a contradiction, since ...


1

Hint: rewrite $$ Z = \sum_{n=1}^\infty X_n\mathbb{1}_{\{N+1 \geq n\}} $$ and apply your "expectation of the sum is the sum of the expectations" idea. Following the comment below, more detail. Write $Y_n = \mathbb{1}_{\{N+1 \geq n\}}$, which is independent of $X_n$. Then for the expectation, you have $$\mathbb{E}[Z] = \sum_{n=1}^\infty ...


-1

In my experience: There seem to be two times here. $t=0$ and $t=T$. A portfolio is a collection of financial instruments. For instance, I could have a portfolio consisting of 3 stocks and 1 bond. Its value today is the sum of the individual values of the instruments today. $V(0)$ is the value of the portfolio at time 0 (today?) $V(T)$ is the value of the ...


2

I agree that @BCLC is right on saying that I have used risk neutral information The Edited Answer is $ V(0) = 15\times90+ 10\times25 = 1600$ Now compute V(T) $$V(T) = 1800, \text{ if stock goes up}$$ $$1800 = 30\times 10 + 100\times 15$$ $$V(T) = 1700, \text{ if stock goes down}$$ $$1700 = 20\times 10 + 100\times 15$$ $V(T) = 15\times A(T) + ...


3

By the Borel-Cantelli lemma, if the series $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$ converges for each $\varepsilon>0$, $Y_n\to0$ almost surely as $n\to\infty$. Using Chebyshev's inequality, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$ since $\operatorname EY_n=0$. By ...


2

Hint: use the Borel—Cantelli lemma. In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$ In even more detail: (place your ...


0

If f is convex, $f(\alpha a + \beta b) \le \alpha f(a) + \beta f(b)$ where two nonnegative coefficents $\alpha \ge 0$, and $\beta \ge 0$ $$\alpha + \beta = 1$$ Now let $\alpha = 0.5*(1 - Y_i/c_i)$ and $\beta = 0.5*(1 + Y_i/c_i)$ and $ a = -kc_i$, and $b = kc_i$ and $f = e^x$ $$ e^{kYi} = e^{(1-Y_i/c_i)/2 (-kc_i) + (1+Y_i/c_i)/2 (kc_i)}$$ $$\le \alpha ...


0

For the inverse transform method, we need an expression for the cumulative distribution function $$ F(x)=P(X<x)=\int_{-\infty}^xf(t)\,\mathrm dt.$$ If $x\le -1$, clearly $F(x)=0$. If $-1\le x\le 2$, $$\begin{align}F(x)&=\int_{-1}^xa(3-t)\,\mathrm dt\\& =\left[3at-\frac12at^2\right]_{-1}^x\\&=3ax-\frac a2x^2+\frac 72a.\end{align}$$ If $2\le ...


0

If there are no additional assumptions, then the smallest possible value of the variance of their mean is $0$ (suppose that both of the random variables are equal to the same constant almost surely). We have that $$ ...


0

There are many ways in which the assumption of IID variables can be broken in real-life applications. Let me illustrate how it can break down in your case of hypertension. I'm not a medical doctor or epidemiologist, so I'll make up some assumptions that are at least plausible to a non-medical doctor. Suppose people who work hard and are financially ...


0

I guess that we can agree on the premises that hypertensive status affected or at least correlated with factors like age, genetics, diet, sport habbits etc. Assume that you have a clinic and you are measuring BP to each person that comes to your clinic (from whatever reason). Assuming, that the measured values are absolutely independent is unrealistic. As ...


0

Using the fact that the density must integrate to one over $[0,1]$, then proportional means $$1 = \int_0^1cf(x)\,dx = \int_0^1\frac{c}{\sqrt x}\, dx = c\left[2\sqrt x\right]_0^1 = 2c.$$ Thus $c = 1/2$. If you now try to compute the expectation, you will find $$\int_0^1x\cdot \frac{1/2}{\sqrt x} = \frac{1}{3},$$ which is option 3.


1

You've neglected the possibility of ties and are over counting events where multiple dice equal $y$. You wish to calculate the probability that two dice are $x$ and $y$ and none of the remaining die are higher than $y$. There are two cases to consider. When $x=y$ and when $x>y$ When $x=y$ you want the probability that all dice are at most $x$, ...


0

$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Corr}{\operatorname{Corr}}\newcommand{\Cov}{\operatorname{Cov}}$ Using the known properties of covariance gives \begin{align*} \Corr(X,Z) &=\frac{\Cov(X,Z)}{\sqrt{\Var(X)\Var(Z)}}\\ &=\frac{\Cov(-10Y+10,Z)}{\sqrt{\Var(-10Y+10)\Var(z)}}\\ &=\frac{-10\Cov(Y,Z)}{\sqrt{100\Var(Y)\Var(Z)}}\\ ...


0

If in doubt, compute. We have $$\text{Cov}(X,Z)=\text{Cov}(-10Y+10,Z)=-10\text{Cov}(Y,Z).$$ Also, $\text{Var}(X)=100\text{Var}(Y)$. It follows that $$\rho(X,Z)=\frac{\text{Cov}(X,Z)}{\sigma_X\sigma_Z}=\frac{-10\text{Cov}(Y,Z)}{10\sigma_Y\sigma_Z},$$ and therefore $r_1=-r_2$. Informally, scaling by some positive factor does not change the degree of linear ...


1

It is correct, but can be further simplified by considering the order statistics of the sample. That is to say, if $$\boldsymbol x = (x_1, \ldots, x_n)$$ is the sample, then consider $$x_{(1)} = \min_i x_i, \quad x_{(n)} = \max_i x_i,$$ the first and last order statistics which are equal to the smallest and largest observations in the sample, respectively. ...


1

To construct counterexamples: Let $Y\equiv 2$ and $X \sim U(-1,1)$. Then $Var(X)>0=Var(Y)$. Let $Y \equiv -2$ and $X\sim U(-1,1)$ or simply $X\equiv 0$. Then $E[X]=0>-2=E[Y]$. Let $Y\in\{-2,2\}$ with $P(Y=-2)=p=1-P(Y=2)$ and $p>0$ and $X\equiv 0$. Then $Var(Y)>0=Var(X)$. Let $Y\equiv2$ and $X\sim U(-1,1)$. Then median $Y=2$ and median $X=0$. ...


1

Hint: Think about different variations of what X and Y are allowed to be given our assumptions. In particular, compare cases when $X$ is either identically zero or uniformly distributed, and $Y$ is identically $-2$ or $+2$. See if you can generate counter examples by using these.


2

For (1): show you can have, for instance, $\operatorname{var} Y = 0$ and $\operatorname{var} X >0$, or $\operatorname{var} Y >0$ and $\operatorname{var} X = 0$. For (2): Show you can have $\mathbb{E} X = 0$ and $\mathbb{E} Y > 0$, or $\mathbb{E} X = 0$ and $\mathbb{E} Y < 0$. For (3): You should see the relation with (1). For (4): Show you ...


0

The expectation calculation for independent $a$ and $b$ is incorrect (you can't pass expected value calculations for vectors through in the same way as you do for scalars); however, the final result is the upper bound for the expected value. Intuitively, you will get the highest value when $a$ and $b$ have to be in the same direction, and the stated end ...


1

If $X=Y$ then 1,2,3,4 would be correct. In particular $X-Y=0$ with probability $1$ If $X$ and $Y$ are independent then 1,2,4 are correct but 3 might not be. For example, suppose $X$ takes the value $1$ with probability $0.4$ and the value $0$ with probability $0.6$ But if $X$ and $Y$ have a more complicated relationship then there is little you can say ...


4

To simplify the subscripts, I'm going to consider just the first few letters the monkey types. Add a variable $k$ to all my subscripts if you want to apply the reasoning at an arbitrary point in the string. It is true that $E(X_1) = 26^{-5}$ and also that $E(X_2) = 26^{-5}$. We just have to cycle through the $26^6$ equally-likely possibilities for the first ...


-5

Since this has intrigued some discussion i decided to post (still) another answer incorporating all the comments around expectation and imposing conditions. The answer does not use expectation but conditional probabilities. Assume a (random) string S of length $len(S) = n$ and an i.i.d alphabet of $N_l$ letters/symbols (for example $N_l=26$). The ...


1

This is not an answer to the question, so this may not be strictly speaking quite welcome here, but I think that any putative proof that the question's expectation formula is not valid should account for this data somehow, and so should any alternative expectation formula for this experiment. The formula in the question seems to match the simulation results ...


1

Where in this proof have we imposed that restriction? Your observation that $X_i=1 \implies \sum_{n=1}^4X_{i+n}=0$ restricts $X_{i+1}\cdots X_{i+4}$ is certainly valid. There is no need to take this restriction into consideration, because it is self-imposing:$$X_i=1 \implies \sum_{n=1}^4X_{i-n}=0$$ Edit: $E[X_{i+1}]\cdots E[X_{i+4}]$ are not ...


5

Note that $E[X+Y]=E[X]+E[Y]$ holds in full generality, even if $X$ and $Y$ are not mutually independent. Proofs of linearity of expectation do not assume independence of $X$ and $Y$. Here's one for example. In other words you do not need to impose that restriction.


2

Note that $X_n\gt 0$ surely for each $n$, hence by definition of $g$, we have $g\left(X_n\right)=1$. The random variable $X$ is not specified but we can imagine that this is the limit in probability of the sequence $\left(X_n\right)_{n\geqslant 1}$. Since $X_n\to 0$ in probability and $g(0)=0$, we do not have $g(X_n)\to g(X)$ in probability (the problem ...


0

Here is a start on part of it. Let $X_1$ and $X_2$ be independent random variables, each distributed $Exp(rate = \lambda) = Exp(mean = 1/\lambda),$ and let $V = \min(X_1, X_2).$ Then $$P(V > v) = P(X_1 > v, X_2 > v) = P(X_1 > v)P(X_2 > v) = [e^{-\lambda v}]^2 = e^{-2\lambda v}.$$ Thus $P(V \le v) = 1 - e^{-2\lambda v},$ and $V \sim ...


3

If $M = \min(X_{1}, . . . , X_{n})$, it can be shown that $M$ has an exponential distribution with parameter $n \lambda_0$. So $E(M)=\frac 1 {n \lambda_0}$ Since $T=nM$, it follows that $E(T)=E(nM)=nE(M)=\frac 1 {\lambda_0}$ Details can be found here: Distribution of the minimum of exponential random variables


1

$$ \mathbb{E}[Z^2\mid M] = \mathbb{E}\left[\left(\sum_{k=1}^N Y_k\right)^2\right] = \mathbb{E}\left[\sum_{k=1}^N \sum_{\ell=1}^N Y_k Y_\ell\right] = \sum_{k=1}^N \sum_{\ell=1}^N \mathbb{E}\left[Y_k Y_\ell\right] $$ by linearity of expectation. Now, $Y_k$ and $Y_\ell$ are independent if, and only if, $k\neq \ell$, so $$\begin{align} \mathbb{E}[Z^2\mid M] ...


1

You say "In both cases it seems advisable to bet everything". But your intuition is not correct. In fact, if you bet almost everything and lose, the log of your final amount of money approaches negative infinity, but if you win it approaches a finite value. So betting everything actually gives you infinitely negative expected value! To see where the $ ...


1

Note that $$P(X^n\in C_n(t))=P(p(X^n)\ge 2^{-nt})=P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le t\right)$$ Now, it follows from WLLN, $$\lim_{n\to \infty}P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le H(X)+\epsilon\right)=1\ \forall \epsilon>0$$ Thus, to have $\lim_{n\to \infty}P[X^n\in C_n(t)]\to 1$, we must have $t\ge \epsilon+H(X)\ \forall ...


2

I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite. Notice this: if $Y_n=\frac{S_n}{\sqrt{2n\log\log n}}$, by properties of random walk we know $\mathbb{E}[Y_n]=\frac{\mathbb{E}[S_n]}{\sqrt{2n\log\log n}}=0$ and $$ Var[Y_n]=\frac{Var[S_n]}{2n\log\log n}=\frac{n}{2n\log\log n}=\frac{1}{2\log\log n}\to ...


2

Think of 75% as the probability that the stock goes up, i.e. $0.75$. Then, the trader gains 100 with probability $0.75$, loses 200 with probability $1-0.75=0.25$. On expectation, what is his gain? More formally: let $X$ be the random variable representing his gain. Then, $\mathbb{P}\{X=100\} = 0.75$, and $\mathbb{P}\{X=-100\} = 0.25$. You are asked to ...



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