# Tag Info

4

You cannot use that equality because, by definition, $X$ and $Y$ have to be integer-valued. Instead of that, express the expectation of $X$ (not necessarily an integer) in terms of $\lambda$ and notice that the expectation of $Y$ must be $15\%$ of that - yielding another Poisson distribution (with a different parameter).

4

The variable $X$ is the sample mean and the variable $Y$ is the sample variance times $(n-1)$. So Basu's theorem implies that they are independent. The distribution of $Y$ is $\chi^2_{n-1}$ as the sum of the squares of the $n$ iid normal random variables $X_i-X$, (where $X$ is used and so there are $n-1$ degrees of freedom instead of $n$).

3

When dealing with ceilings, the proper way to proceed is using probability intervals. $$P(Z=z)=P(z-1<\beta X <=z)=P(\frac{z-1}{\beta}<X<=\frac{z}{\beta})=F_X(\frac{z}{\beta})-F_X(\frac{z-1}{\beta})$$ Where $F_X$ is the cumulative distribution of X. Substituting: $$... 3 Draw a picture. We sketch how to find the cdf F(s,t) in other cases. Let T be the triangle where our joint density "lives." It has corners (0,0), (1,1), and (0,1). We will be referring to it several times. Suppose first that t\gt 1. Draw the point P=(s,t). We want the probability of falling in the region below and to the left of P. If s\le ... 3 Since X and Y are independent, then f_{X,Y}(x,y)=f_X(x)f_Y(y). Now: \begin{eqnarray} P(X<Y)&=&1-P(Y \le X)\\ &=&1-\int_1^2\int_1^xf_X(x)f_Y(y)dydx\\ &=&1-\frac{1}{8}\\ &=&\frac{7}{8} \end{eqnarray} Note that, if you don't want to solve the integral, drawing a picture you will note that P(Y \le X) is the area of ... 2 a) Look at a single bit, say 0. It is transmitted as 000. The probability that two or more 0's will be received, and hence the bit will be decoded correctly, is (0.9)^3+\binom{3}{1}(0.9)^2(0.1). Call this probability a. We used a simple case of the binomial distribution. Now suppose we have an 8-bit word to transmit. The probability it will be ... 2 Hint: if n is even, you can group the terms using Gauss summation:$$y_1 + \cdots + y_n = (y_1+y_n) + (y_2+y_{n-1}) + \cdots.$$Can you say anything about these pairs? Try using the symmetry of the inverse CDF: If n is odd, you can first show that the (n+1)/2-th element is zero, and use the same trick as above to establish your result. 2 Your counterexample is almost a good one -- you meant that  G(x) = 0  when x<0, but you wrote x<1. 2 Through the Spitzer identity, it is possible to find some kind of transform of the distribution of M_n. Well, not exactly. The Spitzer identity involves the expressions M^+_n = \max_{0\le k\le n} S_k, where S_0 = 0, S_k = X_1 + \dots + X_k, k\ge 1. So this translates to the positive part of expression you are interested in. But it is possible to ... 2 \begin{eqnarray}F_{T_x|T>x}(a)&=&P(T_x\le a|T>x)\\&=&P(T-x\le a|T>x)\\&=&\frac{P(T-x\le a,T>x)}{P(T>x)}\\&=&\frac{P(x<T\le a+x)}{P(T>x)}\\&=&\frac{F_T(a+x)-F_T(x)}{1-F_t(a)}...(*)\end{eqnarray}. Now, the last has been developed for all a. Since you know that ... 2 {\bf X}=(X_1,\dots, X_n)^\prime has a multivariate normal distribution with \mu_{\bf X}=\mu {\bf 1} and \Sigma_{\bf X}=\sigma^2 I. Here {\bf 1} is the column vector of all 1s, while I is the n\times n identity matrix. Let {\bf e}_1=(1,0,0,\dots,0)^\prime , and let A be the matrix of an orthogonal transformation that takes the vector \bf ... 2 Partial solution here. Y is a quadratic form. In particular, let \mathbf{1} \in \mathbb{R}^n be a vector of ones, and define$$P_\mathbf{1} = \mathbf{1}\left(\mathbf{1}^{T}\mathbf{1}\right)^{-1}\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\text{.}$$Let$$\mathbf{X}=\begin{bmatrix} X_1 \\ X_2 \\ \vdots\\ X_n \end{bmatrix}\text{.}$$... 2 This is not a correct MGF, because M_X(0) always exists and is equal to 1 but here$$M_X(0)=-0(1-e^0)=0\neq1$$2 First moment of the distribution is μ_1=E[X]=α and the first moment of the sample is m_1=\bar{X}. So, set$$μ_1=m_1 \implies α=m_1$$This is the first moment estimator for α. (Note: This method is confusing at the beginnning, because you think, ok so what? But think that m_1 is your sample mean and so it is known, it will be realized when you collect ... 1 Yes, as you noted, when \phi is not invertable we have to modify the transformation function. Such as, for instance, when \phi(x)=x^2, which is a fold mapping two intervals into one (the negative reals and non-negative reals to the non-negative reals), we modify the transformation formula to account for the fact that we have two "inverse" functions: ... 1 Hint. (not meant to be a complete solution) Exponential families are very different from the usual exponential distribution (but of course, the exponential distribution is a special case of a distribution in the exponential family). Distributions that are of an exponential family can be either continuous or discrete. Here, I'm going to prove the claim for ... 1 The time T until the next event has continuous distribution, with density function f_T(t)=\lambda e^{-\lambda t} for t\gt 0, and f_T(t)=0 elsewhere. The distribution of T is called the exponential distribution with parameter \lambda. The expectation E(T) (mean) of T is given by E(T)=\int_0^\infty t\lambda e^{-\lambda t}\,dt. Integration ... 1 Mostly correct. In the case of a=0 then F_Y(t) = \begin{cases} 0 &: t<b\\1 & :t\geq b\end{cases} because, as you reasoned, Y would be a deterministic random variable (a single massive point at b). So the probability of Y being at most t is zero if t<b and one if t\ge b. I don't see why if X is a discrete random ... 1 You can think about it for a bit. Presumably theres a typo in Y, which should be 1 w.p. \beta. Note that Z is zero iff Y=0 or X=0 (which are independent events), so P(Z=0) = P(Y=0) + P(X=0) - P(X=0)P(Y=0). Note that Z=z where z is a positive integer if and only if Y=1 and X=z (which are independent events), so P(Z=z) = P(Y=1) ... 1 If P(X=n) = \dfrac{\lambda^{n}}{n!}e^{-\lambda} when n is a non-negative integer and 0 otherwise, and Z=\alpha X then$$P(Z=z)=\dfrac{\lambda^{z / \alpha}}{(z/\alpha)!}e^{-\lambda}$$when z/\alpha is a non-negative integer and 0 otherwise 1 No distribution has such a moment generating function, for the trivial reason that M_X(0) = 0, which would imply that$$\operatorname{E}[e^{0X}] = \operatorname{E}[1] = 0,$$a contradiction. 1 This is not really an answer, but notes for future reference about the distribution of X = \max\{X_1,X_2\} when X_1, X_2 \sim \mathcal{N}(0,1), independent, identically distributed. As you stated, the cumulative distribution function (CDF) is$$F_X(x) = P(X\le x) = P(X_1\le x\ \cap X_2 \le x) = P(X_1 \le x) \cdot P(X_2 \le x) = \Phi(x)^2, $$with ... 1 First, your MLE calculation can be made much simpler:$$\mathcal L(\lambda \mid \boldsymbol x) = \prod_{i=1}^n \lambda e^{-\lambda x_i} \mathbb 1 (x_i > 0) = \lambda^n e^{-\lambda n \bar x} \mathbb 1 (x_{(1)} > 0),$$where \boldsymbol x = (x_1, \ldots, x_n) is the sample, \bar x is the sample mean, and x_{(1)} = \min_i x_i is the first order ... 1 You've neglected the possibility of ties and are over counting events where multiple dice equal y. You wish to calculate the probability that two dice are x and y and none of the remaining die are higher than y. There are two cases to consider. When x=y and when x>y When x=y you want the probability that all dice are at most x, ... 1 First, some basic calculations. let p be the probability of guessing a card correctly. Then the probability of getting exactly 3 correct is \binom 53 p^3(1-p)^2. If p=.2 this is .0512, if p=.5 this is .3125 Let's say your prior is \theta_0. That is, before you test anything, you estimate that the "ESP probability" is \theta_0. Then, ... 1 The way you have written it, "\theta = probability that the psychic has ESP", \theta essentially is your prior distribution. There are only two possibilities, ESP and not-ESP, so the full statement of the prior is (I write e for ESP and \neg for negation): P(e) = \theta P(\neg e) = 1 - \theta Writing d for the observed data (3 out of 5 ... 1 We have$$F_Y(y)=\Pr(Y\le y)=\Pr\left(\frac{X}{1-X}\le y\right).$$This is \Pr(X\le y(1-X)), which is \Pr(X(1+y)\le y). Finally, for y positive, which is the only interesting part, we have$$F_Y(y)=\Pr\left(X\le \frac{y}{1+y}\right)=\frac{y}{1+y}.$$Elsewhere, we have F_Y(y)=0. For the density function, differentiate. The second problem is ... 1 1) \max(x_1, x_1 + x_2) \le t if either x_1 \le t and x_2 \le 0, or x_1 + x_2 \le t and x_2 \ge 0. It may help to sketch this in the x_1-x_2 plane. Thus if (X_1, X_2) has joint density f(x_1,x_2),$$P(\max(X_1, X_1 + X_2) \le t) = \int_{-\infty}^t dx_1 \int_{-\infty}^{t-x_1} dx_2 f(x_1,x_2)$$If X_1 and X_2 are iid with density f and ... 1 An observation (to replace a previous wrong answer). Note that$$M_2=X_1+X_2^+$$where X_2^+=\max\{0,X_2\}. So,$$E[M_2]=E[X_1]+E[X_2^+]=0+\frac{1}{σ\sqrt{2\pi}}$$Similarly$$M_3=X_1+\max\{0,X_2,X_2+X_3\}=X_1+\left(X_2+\max\{0,X_3\}\right)^+=X_1+\left(X_2+X_3^+\right)^+ with $E[M_3]>E[X_2^+]=E[M_1]$. And two links here and here that might ...

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 ...

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