# Tag Info

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 ...

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If $U_i$ is independent of $X_i$ then $E(Y_i\mid X_i)=E(X_i\mid X_i)+E(U_i\mid X_i)=X_i+E(U_i)$ hence $E(Y_i\mid X_i)=X_i$. In particular, $E(Y_i\mid X_i=1)=1$ and $E(Y_i\mid X_i=0)=0$. (Also, I calculated $P(Y > \frac{1}{2}) = 0.6728p + 0.1932$ from an earlier part of the question). Actually, if $\sigma^2=\frac13$ then $P\left(Y > ... 4 Yes: as soon as at least one of the random variables$X$and$Y$has a density and if$X$and$Y$are independent, the sum$X+Y$has a density. To understand the magic, assume that$(X,Y)$is independent, that$Y$has density$f$and that$X$is purely discrete with$P(X=x_n)=p_n$for every$n$, then$X+Y$has density$g$with $$g(x)=\sum_np_n\,f(x-x_n).$$ ... 3 By "the series$\sum\limits_n X_n$converge almost surely", it is meant that there exists$\Omega'$of probability$1$for which if$\omega\in\Omega'$, then the series of real numbers$\sum\limits_{n=1}^{+\infty}X_n(\omega)$is convergent. Define the event$E_n:=\{X_n\neq a_n\}$. We get by the assumption that$\mathbb P\left(\limsup\limits_nE_n\right)=0$. ... 3 A$k$-dimensional random vector$X=(X_n)_{1\leqslant n\leqslant k}$is standard Gaussian if the entries$X_n$are i.i.d. standard normal. The (Euclidean) length of$X$is$L_k=\sqrt{X_1^2+\cdots+X_k^2}$, its expected length is$\lambda_k=E(L_k)$. Since the distribution of$X$is completely known, one can compute$\lambda_k$in the usual way, that is, by ... 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$E(X^2 | Y) = YE(X^2) = E(Y) = \frac{1}{2}$implies$X \in L^1$2 In general, no. A standard counterexample is to take your probability space to be$\Omega = [0,1]$with Lebesgue measure, and consider the complex-valued random variables$X_n(\omega) = e^{2 \pi i n \omega}$, which are all bounded in absolute value by 1. They are orthonormal in$L^2(\Omega)$, so by Bessel's inequality, they converge weakly to 0 in ... 2 Remember this following result: Theorem A probability$P$on$(\Omega,\mathcal{A})$and a converging sequence of sets$A_n \in \mathcal{A}$with limit$A \in \mathcal{A}$. Then$\lim_{n \to \infty}P(A_n) = P(A)$In particular you have that if$A_n \to \Omega \implies P(A_n) \to 1$and$A_n \to \emptyset \implies P(A_n) \to 0$. A CDF of a probability ... 2 If$(Y_n)_{n\geqslant 1}$is a sequence of real valued random variables which converge in distribution to$Y$and$g\colon\mathbf R\to\mathbf R$is a continuous function, then$(g(Y_n))_{n\geqslant 1}$converges to$g(Y)$in distribution. 2 For$y>0$: $$P\left\{ X=n\mid\lambda=y\right\} =e^{-y}\frac{y^{n}}{n!}$$ and PDF of$\lambda$is$f_{\lambda}\left(y\right)=e^{-y}$. This leads to: $$P\left\{ X=n\right\} =\int P\left\{ X=n\mid\lambda=y\right\} ... 2 The proof of the convergence in distribution of (X_n,Y) to (X,Y) is done here (the use of characteristic functions could make it shorter). Assume that X=h(Y) for some Borel function h. Using the assumption with g:=f\circ h, a bounded Borel-measurable function, we obtain that for each continuous and bounded function f,$$\lim_{n\to ... 2 To show$E[X|\mathcal{G}]$equals some$Y$. You need to check:$Y$is$\mathcal{G}$-measurable;$\int_A Y(\omega)dP(\omega)=\int_A X(\omega)dP(\omega)$for all$A\in\mathcal{G}$. In this case, the claim is$Y=X$works. (1) holds by assumption and (2) holds trivially. 2 For a fixed$t$, the series defining$X(t)$has only one term which can be different from$0$(for the potential index$n$for which$nT\lt t\leqslant nT+T/2$). Therefore, we can switch the sum and the expectation and we find that$\mathbb E[X(t)]=0$for each$t$because$p(t-nT)$is not random and$\mathbb E[A_n]=0$. For the covariance, by the argument ... 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 The calculation of$E(Y)$is correct. If$a$and$b$are constants, then$E(aX+b)=aE(X)+b$. For the variance, use the fact that if$a$andd$b$are constants, then $$\text{Var}(aX+b)=a^2\text{Var}(X).\tag{1}$$ So the variance of$aX+b$is independent of$b$. Intuitively, this is because the variance is a measure of variability, so is unchanged if you ... 1 If$Y=\limsup X_n$, then there are only finitely many$i$such that$X_i<Y+1$. Specifically, let$Y=\limsup X_n$. Let$N$be the random variable which is the largest$N$such that$X_i\geq Y+1$, or$N=0$. Then$\sup X_n <\max(X_1,X_2,\dots,X_N,y+1)$. (As mentioned above, this assumes that the indexes$i$of$X_i$are bounded below and integers. If we ... 1 The function$\omega \mapsto(X_1, X_2, \cdots X_n)$is a measurable function from$\Omega$to$\mathbb{R}^n$. It is, more strongly, also$\sigma(X_1, X_2, \cdots X_n)$-measurable. Since$f$is measurable, the composition$f(X_1, X_2, \cdots X_n)$is$\sigma(X_1, X_2, \cdots X_n)$-measurable. 1 Consider$X_n\equiv 1+\log n$everywhere, and$\lambda=1$. Then every element belongs to LHS but RHS is empty... so this is a counterexample. 1 By the strong law of large numbers for i.i.d. integrable sequences, $$\frac1N\sum_{i=1}^N\log X_i\to\nu\quad\text{almost surely},$$ where$\nu=E(\log X_1)$. This is equivalent to the assertion that $$\lim_{N \to \infty} \left( \prod_{i = 1}^{N} X_{i} \right)^{1/N}=\mathrm e^\nu\quad\text{almost surely},$$ hence the expected value of the LHS exists and is ... 1 Hint: In full generality (if$U$and$V$are integrable), $$E(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}\cdot E(U)+\color{blue}{3}\cdot E(V).$$ If$U$and$V$are independent (and square integrable), $$\mathrm{var}(\color{red}{5}U+\color{blue}{3}V)=\color{red}{5}^2\cdot \mathrm{var}(U)+\color{blue}{3}^2\cdot \mathrm{var}(V).$$ 1 I think you are looking for this wikipedia page. 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 One simple solution is to adjust the number of widgets produced to reflect 100 hours of work. In your example, Employee #10 produced 100 actual widgets in 80 hours, but$100*(5/4)=125$adjusted widgets in 100 hours. Hence the ten employees made a total of$1025$adjusted widgets. Then, divide the compensation by adjusted widgets. \$1M divided by 1025 is ...

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What you have for b is wrong this is because $$E[T_{n}]=E\left[\sum_{i=1}^{n}Y_{i}\right]=\sum_{i=1}^{n}E[Y_{i}]=\sum_{i=1}^{n}\frac{2}{2^{i}}=\sum_{i=1}^{n}\frac{1}{2^{i-1}}=\sum_{i=0}^{n-1}\frac{1}{2^{i}}$$ Now here you also attempted to use fact about a geometric series, but remember n is finite here (it is not approaching infinity like it does in a ...

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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 ...

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Draw a picture. The joint density function "lives" over the part $D$ of the unit square that is below the half-parabola $y=\sqrt{x}$. To find the density function of $y$, we have to "integrate out" $x$. The function $8x^2y$ is the joint density only on $D$, so we have to confine attention to $D$. Note that at the beginning of $D$ we have $y=\sqrt{x}$, ...

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Evidently so. See this link Are functions of independent variables also independent? which addresses independence. It's clear they are identically distributed, isn't it?

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$X,Y$ are independent if and only if $$P\left\{ X\in A\wedge Y\in B\right\} =P\left\{ X\in A\right\} P\left\{ Y\in B\right\}$$ is true for measurable sets $A,B$. If this is the case and $f,g:\mathbb{R}\rightarrow\mathbb{R}$ are measurable functions then: P\left\{ f\left(X\right)\in A\wedge g\left(Y\right)\in B\right\} =P\left\{ X\in ...

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In every exercise about normal random variables, it proves useful to reduce the problem to the study of a small number of independent standard normal random variables. Here $U=-1+Y$ and $V=1+Z$ where $(Y,Z)$ is i.i.d. standard normal and one knows that, for every $(x,y,z)$, $x+yY+zZ$ is normal $N(x,y^2+z^2)$. This should yield that $X$ is $N(0,2)$ and $T$ ...

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It's hard to tell from a small sample like this. I suggest using the same method in Excel to generate about 1000 points. Then plot the results on a histogram to see what general shape the distribution takes. If the histogram is relatively flat, you're probably getting the uniform distribution. If you see peaks or slopes, you aren't.

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