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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 $1$s, 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


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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