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0

@Andre, I am looking at your elegant solution and I have a question around following step: Now we find Pr(Xi=1). Consider the numbers that are in positions i−1, i, and i+1. There are 3! permutations of these numbers, and precisely 2 of these permutations give a local maximum at i. Thus Pr(Xi=1)=23!=13. If position i attains maxima then i-1 and ...


1

Do you know that $$\mathbb{E}[X|A] = \frac{\mathbb{E}[X1_A]}{P(A)}?$$ Some take it as the definition of conditional expectation given an event. With $X=L$ and $A=(L\geq q_\alpha(L))$, you get your answer.


1

I'm not sure about what you mean about "the relation between $g(x,y)$ and $y$", but I am going to answer what I understand of it: "can we infer than $g$ does not depend on $y$?" The answer is no: take $f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ (pdf of a standard Gaussian), and let $$ g(x,y) \stackrel{\rm def}{=} e^{xy-\frac{y^2}{2}}. $$ Then ...


0

Thanks to Tomas' comment, I went back to my studies and generalized the formula. Having a random variable with two parameters $p_1$ and $p_2$, the Bayes estimator of $p_1$ (and $p_2$ in a similar manner) is simply \begin{align*} &\hat{p_1}=E[p_1|x]=\int p_1\frac{f(x|p_1)\pi(p_1)}{f(x)}dp_1=\int p_1\frac{\int f(x|p_1,p_2)dp_2\pi(p_1)}{\int\int ...


5

Assuming at least that $\Bbb E|X_j|<\infty$: Clearly yes if $\sum|a_j|<\infty$; then Chebyschev says that $\sum P(|a_jX_j|>\epsilon)<\infty$ for every $\epsilon>0$. No in general. Assuming $X_j$ is not essentially bounded there exist $a_j\to0$ such that $\sum P(|a_jX_j|>1)=\infty$, so the less trivial half of Borel-Cantelli says that ...


0

Hint: The sum of independent normally distributed random variables is a normally distributed random variable with (1) its mean being the sum of the means, (2) its variance being the sum of the variances. You want to find $n$ such that $\mathsf P((\sum_{k=1}^n B_k) - C \geq 0)\gt 0.2$, where $\{B_k\}$ are the weights of boxes and C is the capacity of the ...


1

You have $$ (1)=\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}] = \mathbb E[\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}|F_{\tau_a}]] = \mathbb E[1_{\{\tau_a\leq t\}}\Pr(X_{t-\tau_a}\leq 0]|F_{\tau_a})],$$ where you use that $W_{\tau_a}=a$ and measurability of $\tau_a$ wrt. $F_{\tau_a}$. Moreover, ...


9

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


7

Let $X=Z$, and $Y$ such that $P(X \le Y) = \frac{1}{2}$ and $P(X=Y)=0$, then in order for the equality to hold we must have \begin{eqnarray} 0 &=& P(X=Y) \\ &=&P(X \le Y \le Z) \\ &=& P(X \le Y) P(Y \le X)\\ &=& P(X \le Y) \left( 1- P(X<Y) \right) \\ &=& \frac{1}{4} \end{eqnarray} So the equality does not hold in ...


5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ...


2

$$\text{Boy, High income} =4\\ \text{girl, High income} = 6\\ \text{Boy, low income} = 6\\ \text{Girl, low income} = x$$ $$ P(\text{Male - M}) = \frac{10}{16+ x}\\ P(\text{High income - H}) = \frac{10}{16+ x}\\ P(\text{Male High income}- MH) =\frac{4}{16+x} $$ Independence means that $P(M \text{and} H) = P(M) P(H) $ $$\frac{10}{16+x}\frac{10}{16+x} = ...


0

If $X$ is a nonnegative random variable then by definition: $\mathbb{E}X:=\sup\left\{ \mathbb{E}Y\mid Y\text{ is rv with finite range and suffices: }0\leq Y\left(\omega\right)\leq X\left(\omega\right)\text{ for each }\omega\in\Omega\right\} $ If $Y$ is a random variable as described above then it can be written as: $$Y=\sum_{k=1}^{n}r_{k}1_{A_{k}}$$ where ...


2

$$X\leq Y\implies \mathbb E[X]\leq \mathbb E[Y],$$ and thus $$X\geq 0\implies \mathbb E[X]\geq \mathbb E[0]=0.$$


2

I do not know if this is what you are after ... there is another answer given too, but if you know the mean and the covariance of $X$ then: using your comment ($A=B^TB$) and assuming $EX=\mu$ and $cov(X)=\Sigma$ we may write \begin{align} EX^TAX&=E(X-\mu)^TA(X-\mu)+\mu^TA\mu\\ &=E[B(X-\mu)]^TB(X-\mu)+\mu^TA\mu\\ &=\mbox{tr} B\Sigma ...


3

It involves covariance between the various $x_i$. $xx^T$ is a $n\times n$ matrix. Let $C=E\{xx^T\}$. The answer is the trace of $AC$. If $E\{x_ix_y\}=E\{x_i\}E\{x_j\}$, so the $x_i$ are independent of each other, then the answer is $E\{x\}^TAE\{x\}$


1

The characteristic polynomial of $A$ is $$ p_A = t^2-4X t + (5X^2-X) $$ The eigenvalues of $A$ are $$ \lambda_{1,2} = 2X \pm \sqrt{ 4X^2 - 5X^2+X} = 2X\pm \sqrt{X(1-X)} $$ There expected values are $$ \mathbb E[\lambda_{1,2}] = 1 \pm \mathbb E[\sqrt{X(1-X)}]. $$ This is equal to the eigenvalues you found only if $E[\sqrt{X(1-X)}] = \frac12$. If $X$ is ...


1

No, you have to find the random eigenvalues of $A$ and then find its mean. For example here $A$ has the characteristic equation $$\lambda^2-4\lambda X+5X^2-X=0\implies \lambda=2X\pm \sqrt{X-X^2}$$ To find mean eigenvalue of $\lambda$ you need to find expectation of this quantity, which is obviously not going to be the answer you found, in general.


0

Let $f\geq 0$ be a measurable function on $X$ with $\int_X f = 0$. We claim that $f$ is zero a.e. Suppose that this is not the case. For each $\epsilon > 0$, the set $U_\epsilon = f^{-1}(\epsilon, \infty)$ is measurable. Since $f^{-1}(0, \infty) = \bigcup_{n=1}^\infty U_{1/n}$, some $U_\epsilon$ must have measure $\mu > 0$. Thus $\int_X f \geq ...


1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?


1

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ...


2

$$\begin{align} \mathsf P(X < Y) &=\int_1^3 \left\{\int_0^y f_X(x) \operatorname dx\right\}g_Y(y) \operatorname dy\\ &\neq \tfrac{1}{4} \int_1^3 \int_0^y \operatorname dx \operatorname dy \end{align}$$ Here's the problem.   The inner integral's upper bound should be $\min(2, y)$ because the support for $X$ is $(0;2)$.   Watch out for ...


1

Split the inteval and use Baye's Theorem to get $$P(X<Y) = P(X<Y | X<1) P(X<1) + P(X<Y|X\geq1)P(X\geq 1))$$ sind $X$ is uniform on $(0,2)$, we know that $P(X<1)=P(X>1) = \frac12$. Furthermore, $P(X<Y|X<1) = 1$ since $Y$ is uniform on $(1,3)$. This yields $$P(X<Y) = 1 \cdot \frac12 + P(X<Y|X\geq1)\frac12$$ Now $P(X<Y| ...


1

If you want to solve the problem using integrals then you should notice that you have wrong upper limit in the inner integral. It should be min(y,2).


11

This sketch might help. You want the red area as a proportion of the red and blue areas.


1

We .can draw the rectangle and it's interior $ 0 \leq x \leq 2$ and $1 \leq y \leq 3$. Then we can draw the line $y=x$. Let's look at our event. So we should draw the line $y=x$. Now, the region delimited is given by the triangle whose vertices are $(1,1)$, $(2,2)$ and $(2,1)$.The probanility is $\int\limits_{1}^{2}\int\limits_{1}^{x} \frac{1}{4}dydx$.


0

Answer: Divide the regions of X with respect to Y for the condition $X<Y$. For $0<X<1$, $P(X<Y) = \frac{1}{2}$ For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$ For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$ Thus $P(X<Y) = ...


4

The simplest way is probably using the CDF, which is known in this case. Let $X \sim Exp(1)$ and $Y = \sqrt{X}.$ The CDF of $X$ is $F_X(x) = 1 - e^{-x}$, for $x > 0.$ Then $$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = 1 - e^{-y^2},$$ for $y > 0.$ Then take the derivative of the CDF of $Y$ to get $F_Y^\prime(y) = f_Y(y) = 2ye^{-y^2},$ ...


2

$$M(t)=\frac{1}{100}e^{-2t}+\frac{3}{20}e^{-t}+\frac{237}{400}+\frac{9}{40}e^{t}+\frac{9}{400}e^{-2t}\\=\frac{e^{-2t}}{400}\left(4+60e^{t}+237e^{2t}+90e^{3t}+9e^{4t}\right)\\=\frac{e^{-2t}}{400}\left(2+15e^{t}+3e^{2t}\right)^{2}$$ So $$M_{X}(t)=\frac{1}{20}(2e^{-t}+15+3e^{t})$$ And by definition, $$M_{X}(t)=\mathbb{E}(e^{tX})=\sum p_{i}e^{x_{i}t}$$ So ...


2

By the definition of $T$ we have $Z_T>2\alpha$ whenever $T<\infty$. If, in addition, $|Z_n-Z_T|\leq \alpha$ we get $$|Z_n|-\alpha=\alpha + |Z_n|-2\alpha > \alpha + |Z_n|-|Z_T| \geq \alpha - |Z_n-Z_T|\geq 0.$$


1

This is merely a slight extension of Nate's comment, but it should answer your question: On $(\Omega,\mathcal{A},\mathbf{P})$ we have a sequence $(X_i)_{i\in \mathbb{N}}$ of independent RVs $X_i:(\Omega,\mathcal{A})\to(\Omega_i,\mathcal{A}_i)$ and the corresponding sequence $\sigma(X_i)_{i\in \mathbb{N}}\subset\mathcal{A}$ of independent $\sigma$-algebras, ...


1

Here is an example of how you can make your idea rigorous. Let $X_1,X_2,X_3,\dots$ be a sequence of independent real random variables. Let us fix some integer $k$ and let $$ \mathcal F = \sigma(X_m ; m \leq k), \qquad \mathcal G = \sigma(X_m ; m \geq k+1). $$ For all $n \geq k+1$, the random variable $Y_n = \sup_{m \geq n} X_m$ is $\mathcal G$-measurable. ...


3

Let $\Omega = \{a,b\}$. Define $X,Y:\Omega\to\mathbb R$ by $$X(\omega) = \begin{cases}0,& \omega=a\\ 1,&\omega=b. \end{cases} $$ and $$Y(\omega) = \begin{cases}1,& \omega=a\\ 0,&\omega=b. \end{cases} $$ Then $\sigma(X)=\sigma(Y) = 2^\Omega$, but $XY$ is identically zero so $\sigma(XY)=\{\varnothing,\Omega\}$.


2

Let $\mathcal A,\mathcal B,\mathcal C$ be three independent $\sigma$-algebras and let $\mathcal{D}$ be the $\sigma$-algebra generated by $\mathcal{B}$ and $\mathcal{C}$. We want to show that $\mathcal{A}$ and $\mathcal{D}$ are independent. Indeed: if $\mathcal A,\mathcal B,\mathcal C$ are the natural $\sigma$-algebras of $X,Y,Z$ the product $YZ$ is a ...


3

[This answers the original edition of the question, which did not assume $X_1,X_2,X_3,\ldots$ are independent.] No. Suppose $R\sim\mathrm{Uniform}(0,1)$ and $(Y_1,Y_2,Y_3,\ldots)\mid R\sim\mathrm{i.i.d. Bernoulli}(R)$. Then the strong law of large numbers implies that $$ \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) = 1. $$ So ...


3

It sounds like you want to generate gamma-distributed random variables. If you want to write the code yourself, the way I know how is using the acceptance-rejection method by first generating an exponentially-distributed RV. An exponential random variable $G$ with pdf $\lambda \mathrm e^{-\lambda g}$ may be generated by first generating a uniform(0,1) ...


0

If $F$ is the pdf of the Gamma distribution with the correct parameters then for all real $x$, $$\Pr[\Delta X < x] = F(x).$$ This defines the distribution of $\Delta X$ completely.


1

Counterexample: Say, $\sigma=1$. Take $X_N$ to be the random variable taking values in $\left\{-\frac{1}{N},N\right\}$ for $N>1$ with $\text{Prob}\left(X_N=-\frac{1}{N}\right)=\frac{N^2}{N^2+1}$ and $\text{Prob}\left(X_N=N\right)=\frac{1}{N^2+1}$. Obviously, $E\left[X_N\right]=0$ and $\text{Var}\left(X_N\right)=1=\sigma^2$. Now, ...


1

Your computation (actually, estimation) of the entropy in terms of a sequence, implicitly assumes that the succesive symbols are independent (and the source is stationary, of course). However, it seems intuitively true that the second sequence is more "random" than the first sequence and hence should, in some sense, have higher entropy. Your ...


0

Let $\{ X_n \}_{n\geq 1}$ be a sequence of independent random variables such that $X_n=1$ with probability $1/n$ and $X_n=0$ otherwise. Then $$ \mathbb P(X_n > 0)=1/n\rightarrow 0 $$ and we have convergence in probability to $0$. However, for any small $\epsilon > 0$ $$ \sum^\infty_{n=1} \mathbb P(X_n>\epsilon) = \sum^\infty_{n=1} \mathbb P(X_n=1) ...


0

To be found is the integral $$\int\int f\left(x,y\right)\left[y>x\right]dxdy$$ where $\left[y>x\right]=1$ if $y>x$ and $\left[y>x\right]=0$ otherwise. This comes to the same as finding $$\frac{3}{5}\int_{0}^{1}\int_{0}^{y}x\left(y+y^2\right)dxdy$$ Can you take it from here?


0

Draw a $2 \times 1$ rectangle in the $xy$-axis and integrate the cdf in the region of the rectangle where $Y>X$. Note this region can be described in terms of the line $Y=X$.


0

My answer may be not rigorous but it's simple. I derive the joint density without cumulative function as follows: Let $\tau_{j}$ denote a permutation of $1,\dots,n$, and each permutation do not equals to each other. Thus, we have $$ \begin{align*} & f_{X_{(1)}, X_{(2)}, \dots, X_{(n)}}(y_1, \dots, y_n) \\ = & \Pr \{X_{\tau_1(1)} =y_1, \dots, ...


0

I figured out something that worked for what I needed, which was slightly different from the question I'd originally posed. It was based on this inequality: $$ e^x-1 \leq |x|\,e^x. $$ Thus, $$ e^X - 1 \leq |X|\,e^X, $$ almost surely, which implies $$ E[e^X-1] \leq E[|X|\,e^X]. $$ With $E[X]=0$ and $e^x$ convex, Jensen's inequality implies $$ 1 = e^0 = ...


1

You should have a feeling that this couldn't possibly be true. Ignoring the mean zero condition, consider $X$ a constant. The left side increases exponentially, but the right quadratically. Now considering the mean zero condition again, if we just take $X=+x$ and $X=-x$ with both probabilities $1/2$, then $E[(e^{X/2}-1)^2]$ still scales exponentially, ...


3

We have $X,Y$ real valued with $Y(\omega)\leq X(\omega)\Leftrightarrow0\leq X-Y$ for all $\omega\in \Omega$ and $\mathbf{E}(X),\mathbf{E}(Y)$ exists, we set as mentioned in the comments $$ Z=X-Y\geq0 $$ Now we just plug in and get $$ ...


1

For your information, here is the relevant quote from page 27 of A Course in Probability Theory (2nd edition, Academic Press, 1974) by Kai Lai Chung. He has just finished constructing the probability measure $\mu$ with distribution function $F$. There is one more question: besides the $\mu$ discussed above is there any other p.m. $\nu$ that ...


1

Since the discussion got a little detailed, I'll just write out an enumeration. Only way to be sure. The "one bored guy" states: one T: STSSS, SSTSS, SSSTS 3 in total, each with probability $p(1-p)^4$ two T: TSSTS TSTSS STSST SSTST 4 in total each with probability $p^2(1-p)^3$ three T: TSTST, TTSTS, STSTT 3 in total each with probability ...


0

Yes. \begin{align} \text{Var} \left [ \frac{\sum_{k=2}^{n} X(k)}{n - 1} \right ] &= \frac{1}{(n - 1)^2} \sum_{k=2}^{n} \text{E}[X(k)^2] \\ &= \frac{2}{n - 1} \end{align} since $\text{E}[X(k)^2] = 2$ for every $k$. Hence $\sum_{k=2}^{n} X(k) / (n - 1) \stackrel{\text{p}}{\to} 0$ by Chevyshev's inequality.


0

Lets start with the bivariate case: Let $Z=X_1X_2$, then the density of $Z$ is: $$f_Z(z)=\int_{-\infty}^{\infty} f_X(y)f_X(z/y) dy = \frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(-\frac{y^2+\frac{z^2}{y^2}}{2}\right)dy$$ Now, since the $x_i$ are independent, we can multiply $Z$ by $X_3$ to get $Z_2$ our next iteration: Going beyond this is a bit tricky. ...


0

Update: By Lebesgue-Stieltjes-Correspondence, there is a one to one correspondence between the probability law $P_X$ when $X$ is Borel measurable and the c.d.f $F_X$. I guess what the text meant is that we could have a random variable $Y$ being measurable to a richer sigma algebra, say, Lebesgue measurable sets than the Borel sets so that the correspondence ...



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