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Clearly, when $b=0$, $v=a$, so it is not a uniform random variable. When $b \neq 0$, we have that $b$ has an inverse in $\mathbb{Z}_p$, so we have for $c \in \mathbb{Z}_p$ $$\mathbb{P} ( v = c) = \mathbb{P} ( a + b r =c) = \mathbb{P} ( b r = c - a) = \mathbb{P}(r =b^{-1}(c-a)) .$$ Since $r$ is uniform, we have $\mathbb{P}( r = \hat{c} ) = \frac{1}{P}$ for ...

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Hint: $\max(x_1, \ldots, x_n) \le x$ is equivalent to $(x_1 \le x) \wedge (x_2 \le x) \wedge \ldots \wedge (x_n \le x)$.

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Hints: You integrate the joint density $f(x,y)$ with respect to $y$ to get the marginal of $X$ i.e. $f_X(x)$. Integration must be done with proper limits. Similarly you integrate the joint density $f(x,y)$ with respect to $x$ to get the marginal of $y$ i.e. $f_Y(y)$. Here again note the proper limits of integration. Now to show that $X$ and $Y$ are ...

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The random variable $V$ is log-normally distributed if and only if $\ln V$ is a normal random variable. Since the logarithm is a monotonous function, the median of $\ln V$ is the logarithm of the median of $V$. Equivalently, the median of $V$ is the exponential of the median of $\ln V$. (1) The variable $\ln V$ has a normal distribution, which is ...

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To complete the proof you need to show that $\lambda_n\rightarrow \lambda$ implies $(1+\lambda_n/n)^n\rightarrow e^{\lambda}$. Here's the proof: $$(1+\lambda_n/n)^n=\exp(n\log(1+\lambda_n/n)).$$ Now expand the log: $$n\log(1+\lambda_n/n)=n\left(\frac{\lambda_n}{n}+O(\lambda^2_n/n^2)\right)=\lambda_n+nO(\lambda^2_n/n^2).$$ Notice that $n\cdot ... 1 (a) You are correct,$Y$can be either$0,1,2$or$3$. (b) You simply need to calculate$P(Y=0), P(Y=1), P(Y=2)$and$P(Y=3)$. How would you calculate$P(Y=0)$? Remember,$P(Y=0)$is the probability that no errors occured, which means that it is the probability that no error occured in the first transmission and no error occured in the second transmission ... 1 A random Gaussian process$v = (v_k)$with a covariance matrix$U$can be represented by$v = U^{1/2} g$, where$g$is a vector of i.i.d.$\mathcal N(0,1)$random variables. So it would seem reasonable that a$(n \times p)matrix is called "distributed according to a matrix valued normal distribution" if it has some kind of representation like: $$X = ... 1 Let's have three biased coins. Coin A produces heads with probability \alpha, coin B produces heads with probability p and coin C produces heads with probability q. Here is the experiment: We flip coin A. If the result is heads then we flip coin B otherwise we flip coin C. The result of the experiment is the result of the second flip. We ... 1 The pair (X,Y) is uniformly distributed in a triangle. That does not imply that either X or Y is uniformly distributed. Draw the triangle, and you'll see why X is more likely to be in a short interval near 1 than in an interval of the same length near 0, so X will not be uniformly distributed. Similarly, Y is more likely to be in a short ... 1 Imagine the selected elements as black beads and the others as white. Add one more black bead (so there are b+1 black beads and a+1 beads in all) and close the row of beads into a circle. The circle is divided into b+1 segments, each consisting of some (possibly zero) white beads followed by one black bead. By symmetry, each of these segments has ... 1 Start out with the b selected elements in a row, and then insert the remaining a-b elements one by one, with equal probability for every possible insertion point (including the two ends). This gives you all possible arrangements of the selected elements with equal probability (since each arrangement is obtained in (a-b)! different ways). In the ... 1 The expected value is a sort of average value over a large # of trials. Suppose an elementary event has a Pr = \dfrac{1}{100}, i.e. 1 in 100 On an average, how many trials would be needed to get the event once ? 100, isn't it, as the term itself suggests ? If an elementary event has a probability p, E[x] = \dfrac{1}{p} More complex computations are ... 1 Another approach is to use the Law of Iterated Expectation, and partitioning on the event of succeeding on the first (next) try. Let \bar X be the expectation of the count of tries until you encounter a six. If you succeed on the first try, the (conditional) expectation is 1; this condition has probability \tfrac 1 6 of occurring. If you ... 1 This is because you're computing a probability, not an expected value. There is no need to introduce two variables. Let X the number of times you roll the die before you get a 6. You have P(X=k)=p(1-p)^k, for k\geq 0, with p=\frac{1}{6} (by the way this is a geometric distribution). If you want the expected value, then you have to compute: ... 1 I know the marginal distribution to be the probability distribution of a subset of values, Yes. In this case, the subsets of \{X, Y\} we're interested in are \{X\} and \{Y\}. You have been given the joint density function, f_{X,Y}(x,y), and the support for this function, 0\leq x\leq 2, 0\leq y\leq 1. To obtain the marginal density ... 1 Hint:$$\begin{align*} f_X(x) &= \int\limits_{-\infty}^{\infty}f_{X,Y}(x,y)\text{ d}y \sim \int_{\text{outside of }[0, 2]}0+\int\limits_{0}^{2}\dfrac{3}{2}y^2\text{ d}y = \int\limits_{0}^{2}\dfrac{3}{2}y^2\text{ d}y\\ f_Y(y) &= \int\limits_{-\infty}^{\infty}f_{X,Y}(x,y)\text{ d}x \sim \int_{\text{outside of }[0, ... 1 LetX_i \sim Pois(\lambda_1)$be the number of arrivals in one hour, in effect, we have observations$X_1, X_2, \dots, X_{100}.$Because$\bar X = 2$our point estimate of the Poisson hourly rate of arrival is$\hat \lambda_1 = 2.$We do not know the individual$X_i$but we do know know$T = \sum_{i=1}^{100} X_i = 200.$Because the sum of Poissons is ... 1 In my opinion, it is easier to start by the cdf. Let$z\in\mathbb{R}$.$P(Z\leq z)=P(|Y-\delta |\leq z)=P(\delta - z\leq Y\leq \delta+z)=F_Y(\delta+z)-F_Y(\delta -z)$where$F_Y(y)=(1-\exp(-\Omega y))\mathbf{1}_{y\geq 0}$is the cdf of$Y$. 1 If the sample space of a random variable is finite then also the$\sigma$-algebra$\mathcal A$on it is finite. Then set$\{P(A)\mid A\in\mathcal A\}$must be finite as well. However, this set is infinite if you are dealing with a random variable that has geometric distribution having parameter$p\in(0,1)$. In that case for nonnegative integers$k,m$we ... 1 due to symmetry, half the women are shorter that 165, so you need to calculate $$p(M<165)=p\left(z<\frac{165-178}{8}\right)$$ Can you finish it? 1 I would break this question up into two pieces. First, what height are half of the women taller than? Since the mean of the women distribution is$165$and the median of a normal distribution is its mean, the answer is 165. The second half of the question is: What proportion of men are less than the height we just calculated. i.e. What proportion of men ... 1 As you outlined, the statement that$X^- =^d Y^-$means that $$P[X \leq s] = P[Y \leq s]$$ for all$s \leq 0$. So we want to show:$X =^d Y$, i.e.$P[X \leq u] = P[Y \leq u]$for$u \in \mathbb{R}$. If$u \leq 0$we are done by the above result. Suppose instead that$u > 0\$. Then $$P[X \leq u] = 1 - P[X > u] = 1 - P[Y > u] = P[Y \leq u]$$ Here ...

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