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1

Suppose $Y$ is exponentially distributed with expected value $\sigma$ so that $\Pr(Y>y)=e^{-y/\sigma}$, and let \begin{align} Z_1 & = \sqrt{-2\log_e Y}\,\cos X, \\ Z_2 & =\sqrt{-2\log_e Y}\,\sin X. \end{align} Then each of $Z_1,Z_2$ is normally distributed with expected value $0$ and variance $\sigma^2$ and $Z_1,Z_2$ are indepedent. See this ...


0

We have: $$\begin{eqnarray*}\mathbb{P}\left[Y=\frac{Z}{\cos(X)}\in[0,\eta]\right]&=&\frac{1}{\sqrt{2}\,\sigma\,\pi^{3/2}}\int_{0}^{\eta}\int_{-1}^{1}\frac{e^{-\frac{(ux)^2}{2\sigma^2}}}{\sqrt{1-x^2}}\,dx\,du\\&=&\frac{1}{\sigma\sqrt{2\pi}}\int_{0}^{\eta}e^{-\frac{u^2}{4\sigma^2}}\,I_0\left(\frac{u^2}{4\sigma^2}\right)\,du\end{eqnarray*}$$ ...


1

Since $\mu$ is a parameter and not a random variable, it having a covariance does not make sense. Even if you consider it as random variable, apart from the above assumptions, you have to define its probability distribution. Covariance is, like the mean value, an integral with respect to the probability distribution of the random variables. Here, you have ...


0

Your method for part a is correct. For part b, realize that $$ \log Y = \frac1n\sum_{i=1}^n -\log(U_n) $$ We can now use the central limit theorem on $\log Y$, since $\log Y$ is an average of independent random variables each with the same distribution, namely, the distribution of $X$. Letting $\mu,\sigma^2$ be the mean and variance of $X$, the Central ...


0

$\left\{ Z>0\right\} =\cup_{n=1}^{\infty}\left\{ Z>n^{-1}\right\} $ and consequently $\sum_{n=1}^{\infty}P\left\{ Z>n^{-1}\right\} \geq P\left\{ Z>0\right\} >0$. Then $P\left\{ Z>n^{-1}\right\} >0$ must be true for some $n$. Taking $b>0$ small enough (e.g. $b=\frac{1}{2}P\left\{ Z>n^{-1}\right\} $) we also have $P\left\{ ...


1

Suppose otherwise, that $P(Z>a)=0$ for all $a>0$. If your cdf is continuous, then $$0=\lim_{a \to 0}P(Z>a)=P(Z>0) \ne 0,$$ a contradiction.


0

Sometimes one uses the notation $\operatorname{supp}(X)$, the support of $X$, or the support of the probability distribution of $X$, to mean the closure of the set of all points in $\mathbb R$ for whose every open neighborhood the probability that $X$ is in that neighborhood is positive. For a discrete distribution, that is simply the set of all values that ...


0

Since you assume $t$ to be a natural number and properties of $\cos$ it can be shown that $Z_t$ and $Z_{1}$ are equivalent. For $Z_2$, you practically have two $Z_1$ and you toss a coin which one of them to select. Anyway, you have $Z_1$ at the end. Similartly for $t>2$.


1

Use LIE three times, then simply because of independence of the two coordinates w.r.t. each other. $\begin{align} \mathsf E[Z] & = \mathsf E_{X_1}\left[\mathsf E_{X_2\mid X_1}\left[\mathsf E_{Y_1\mid X_1,X_2}\left[\mathsf E_{Y_2\mid X_1,X_2,Y_1}\left[\frac{Y_1+Y_2}{X_1+X_2}\middle| X_1,X_2,Y_1\right]\middle| X_1,X_2\right]\middle| X_1\right]\right] ...


0

Hint: Don't split the numerator. Use both $E[Y_1|X_1] = \theta X_1$ and $E[Y_2|X_2] = \theta X_2$.


0

Let $f$ be the p.d.f. of $X$ (exponential distribution) and let $g$ be the p.d.f. of $Y$ (beta distribution). The c.d.f. of $W=XY$ is then: $$\int_0^1 \int_0^{w/y} f(x)g(y) \, dx \, dy$$ The bounds of integration come from the support of $X$ and $Y$ (i.e. X has a exponential distribution so $x \ge 0$, and $Y$ has a beta distribution so $0 \le y \le 1$).


1

Let $Y=f(U)$. You know that $E(Y)=E(f(U))=\frac{1}{b-a}\int_{a}^{b}f(u) du$. You also know (no?) that $\sum f(U_i)/n =\sum Y_i/n \to E(Y)$ in probability (law of large numbers) if the variance of $Y$ is finite (sufficient condition). Now, $Y$ is bounded (because a continous function on a closed interval is bounded). Hence, it's variance is finite, and ...


0

to have $A,B$ such that: $A-B \sim U$, both $A,B$ must be linearly trasformed samples of the same uniform distribution: mini-proof: $$ \mbox{let: }u \sim U[0,1]\\ \begin{cases} s_1\leftarrow u\cdot \lambda_1+\mu_1\\ s_2\leftarrow u\cdot \lambda_2+\mu_2 \end{cases} \implies \\ s_1-s_2=u\lambda_1+\mu_1-(u \lambda_2+\mu_2) = ...


1

Dejan's answer is absolutely correct, let me elaborate a little bit, starting with the more basic question: 'What is a function?' A function (from the set of real numbers with values in the set of real numbers), is a rule, which assigns to each real number another real number. (For simplicity, I am assuming that the domain of the function is the entire ...


1

When you write $f(x)=x^2$ you usually want to say something like "$f(x)$ is equal to $x^2$". Here, $x$ might be a particular number, like $3$, or $x$ might stand for any possible number, i.e. the meaning of "for all $x$, $f(x)$ is equal to $x^2$" might be intended. Instead of just "$f(x)$ is equal to $x^2$" it might also mean "$f(x)$ is defined as $x^2$". ...


1

For every nonnegative integer $n$, the density $f$ defined on $x\gt0$ by $f(x)=\frac1{n!}x^n\mathrm e^{-x}$ is the PDF of the gamma distribution $(n+1,1)$. Your case is when $n=3$. The parameter $n$ could be any real number $n\gt-1$ provided $n!$ is replaced by $\Gamma(n+1)$.


2

$\{n:n>0,X_n=X_0\}$ is the set of non-zero indices ($n$) of random values ($X_n$) which are equal to $X_0$. The minimum value is the least such index; the first in the list. $\begin{align} \mathsf E[N] & = \mathsf E[\min\{n:n>0, X_n=X_0\}] \\ & = \sum_{n=1}^\infty n \mathsf P(n=\min\{n:n>0, X_n=X_0\}) \\ & = \sum_{n=1}^\infty n \mathsf ...


3

What does $N=\min\{n>0:X_n=X_0\}$ even mean? The set $S=\{n\gt0:X_n=X_0\}\subseteq\mathbb N$ is random, its minimum $N$ is a random variable. For example, if $(X_0(\omega),X_1(\omega),X_2(\omega),X_3(\omega),X_4(\omega))=(7,2,42,7,13)$ then $N(\omega)=3$.


0

Here is an approach from scratch without using any probabilistic tools. Surely better bounds are available. Denote your variable by $X$ and let it take values $x_i$ with probabilities $p_i$. Assume the variance $V(X)$ of $X$ is positive and finite, otherwise the question is trivial. Now define $$Z=X-E(X),$$ so that $E(Z)=0$ and $V(Z)=V(X)$. Call ...


0

Since you are asking the max value of $Pr(X=a)$, I assume $X$ is a discrete random variable. If I can assume to know the set of the values the discrete random variable take, then let the alphabet of the discrete random variable be ${\cal A} = \left\{a_1, a_2, \ldots, a_N\right\}$. Now, we would like to find the PDF $p(X)$ such that $p(a_1)$ (w.l.o.g) is ...


2

$E$ stands for expected value and $d(\cdot, \cdot)$ means distortion, however you define distortion. $X^n$ is a vector or $n$-tuple of random variables $X$. You could write this as $(X_1, X_2, \ldots, X_n)$ where the $X_i$ are random variables with a common distribution (but need not be independent). $f_n(X^n)$ is the result of compressing or encoding ...


0

Let us put $I=\inf({\sf Var}(X)|X\in\phi)$ and $S=\sup({\sf Var}(X)|X\in\phi)$. We will compute $I$ and $S$. Trivially $I=0$ (take $X$ constant equal to $\frac{1}{3}$). Let us now compute $S$. To ease notation, we introduce the characteristic functions $\alpha=1_{X\in[0,\frac{1}{4})}$ and $\beta=1_{X\in[\frac{1}{4},1]}$, so that $\alpha+\beta=1$, and we ...


0

Take $x_1 = 1_{[0,{1 \over 2})}$, $x_2 = 1_{[{1 \over 2},1]}$ on $[0,1]$. Then $E[x_1^k] = E[x_2^k] = {1 \over 2}$ for all $k$, but clearly $x_1 \neq x_2$.


0

Let $\varphi(t)=\mathbb{E}[e^{itX}]$. Then: $$i^n\,\mathbb{E}[X^n] = \varphi^{(n)}(0)$$ hence the problem of finding the distribution of $X$ given its moments is equivalent to finding an inverse Fourier transform for the characteristic function $\varphi(t)$.


1

You can't find the random variable itself, but you may be able to find its distribution. Look up "moment generating function" or "characteristic function".


1

a) For $x$ in the interval from $0$ to $1$, we want $\int_0^x 3(1-t)^2\,dt$. One way to evaluate is to make the substitution $1-t=u$. Then $dt=-du$ and after not much manipulation we end up at $$\int_{u=1-x}^{1} 3u^2\,du.$$ The rest of the calculation is straightforward. Alternately, we can expand $3(1-t)^2$, getting $3-6t+3t^2$ and integrate from $0$ to ...


4

For every increasing function $g:[0,+\infty)\to[0,+\infty)$ and every nonnegative $w$, $$[W\leqslant w]=[g(X)\leqslant w]=[X\leqslant g^{-1}(w)],$$ hence $$F_W=F_X\circ g^{-1}.$$ Now, your task is to check that the function $g$ you are considering is increasing.


1

If $X+Y = 2$ and $X-Y = 1$, then we must have $X = 1.5$ and $Y = 0.5$. Is that even a possible outcome? If not, then $\Pr[X+Y = 2 \ \text{AND} X-Y = 1] = 0$. Note: In general, if $X$ and $Y$ are integers, then $X+Y$ and $X-Y$ are either both even or both odd because their difference $(X+Y)-(X-Y) = 2Y$ must be even.


1

The CDF of a discrete random variable $X$ is continuous everywhere except at those discrete points $x_i$ for which $P\{X=x_i\} > 0$. At these points of discontinuity, the limit from the right exists as does the limit from the left, but these two limits have different values. Let $F_X(a^{-}) = \lim_{x\uparrow a} F_X(x)$ denote the limiting value of ...


3

And it is not hard to see that $E[N]=E[\tilde{N}]$... Actually i is not hard to see that $E[\tilde{N}]=E[1+N]$ since, when evaluating $\tilde N$, one assumes that the first toss after getting a first "4" is not a "4" hence one needs a supplementary toss when compared to $N$. After that, everything flows smoothly and one gets $E[N]=42$.


2

Since you're working with a discrete random variable, the probability of X being lower than b equals the probability of X being equal to or lower than the number before b. Imagine we're working with a fair dice: $P(2\leq X < 5) = P(2\leq X \leq 4)$. Sounds quite fair, doesn't it? I think you can answer your question yourself now :)


0

The negation of your implication is $E(X) > E(Y)$ and $E(f(X)) < E(f(Y))$. Intuitively this will happen if $Y$ takes on a large value with low probability, $X$ only takes on moderate values, and $f$ grows fast enough. For example, consider $X=1$ with probability $1$, $Y=0$ with probability $19/20$, $Y=10$ with probability $1/20$, $f(x)=10^x-1$. Then ...


0

Since the random variable $X$ takes values $x$ and $-x$ with the same probability (i.e. the pmf of $X$ is symmetric about 0), intuitively, we can expect that its average value will be 0. This can be proved formally using the formula for the expected value: $E(X) = \sum_x x P(X=x) $ $= \sum_{x < 0} x P(X=x) + \sum_{x>0} xP(X=x)$ $ = \sum_{x >0} ...


0

Alternative for finding $\mathbb E(X)$: $X$ and $-X$ have the same distribution (by definition $X$ is symmetric) so that $\mathbb E(-X)=\mathbb EX$. This can used as follows: $0=X+(-X)\Rightarrow 0=\mathbb E(X+ (-X))= \mathbb EX+\mathbb E(-X)= 2\times\mathbb EX$ hence $\mathbb EX=0$ In general if $X$ is symmetric and $\mathbb E(X)$ is defined then ...


0

I would say the question is not well posed. Indeed, $P(Y \in B | X \in A)$ is itself a random variable. Indeed by definition of conditional expectation there exists a measurable map $f_B$ such that $P(Y \in B | X \in A) = f_B(1_{X \in A})$, i.e. this is a function of the random variable $1_{X \in A}$.


0

Outline: Draw the $1\times 1$ square on which the joint density function "lives." Draw the lines $x-y=0.5$ and $x-y=-0.5$. We want the probability that the pair $(X,Y)$ lands in the part $A$ of the square that lies between these two lines. To do that, we want to integrate the joint density over the region $A$. For me is easier to find first the integrals ...


0

Can you just show that since: $$P(Y_{n+1}|Y_n) \neq P(Y_{n+1}|Y_n,Y_{n-1}) $$ Then $Y_n$ cannot be a markov chain as it violates the markov property


0

Imagine that the chips all have student numbers, to make them distinct. Since all orderings of the chips are equally likely, it is equally likely that the red chip is first, second, third, and so on. So $\Pr(X=k)=\frac{1}{10}$ for $k=1,2,\dots,10$.


3

Hint: Continue to draw chips even after the red chip is drawn. This produces a uniform random ordering of the chips hence, by symmetry, the position $X$ of the red chip in the whole sample is uniformly distributed, that is, $P(X=k)$ does not depend on $k$ in $\{1,2,\ldots,10\}$. Can you finish?


1

You are correct that the probability of 0 heads is equal to the probability of 4 heads -- there is one way to get each, and each coin flip has equal probability so the probability for each case is $$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$$ The probability of getting 1, 2, or 3 heads is different. For example, ...


0

And I also initially thought that the probability of getting 0 heads is just as likely as getting 4 heads given that we use a normal fair coin This is correct. The combination for 1 head and 3 tails is $h,t,t,t$ and gives $\frac{4!}{3!\cdot 1!}$ permutations. Thus the probability for 1 head and 3 tails is $P(X=1)=\frac{4!}{3!\cdot 1!} \cdot ...


3

$P(X=k) = P(X \ge k) - P(X \ge k+1) = p^k - p^{k+1} = p^k (1-p)$


3

HINT: $$P(X=k) = P(X \geq k) - P(X \geq k+1)$$


1

The problem is with inequality manipulation, which is on the tricky side. For the cdf of $Y$, we want $\Pr(X^{-a}\le y)$. We separate out the case $y\le 0$, although there is no need to. If $y\le 0$, we have $\Pr(Y\le y)=0$. Now we deal with positive $y$. We have $X^{-a}\le y$ precisely if $\frac{1}{X^a}\le y$. This is the case if $X^a\ge y^{-1}$, or ...


2

The result that I am obtaining using Maple is $$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k \right) k!}}={\frac {\Gamma \left( a+1 \right) -a\Gamma \left( a,-\lambda \right) }{{{\rm e}^{\lambda}}a \left( -\lambda \right) ^{a}}} $$ or $$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{ \left( a+k ...


2

Hint: write $$\sup_i\mathbb E|X_i|\leqslant \max_{1\leqslant j\leqslant N}\mathbb E|X_i|+\sup_{j\geqslant N}\mathbb E|X_j-X|+\mathbb E|X|.$$ If $N$ is chosen such that $\sup_{j\geqslant N}\mathbb E|X_j-X|\lt 1$, we get $$\sup_i\mathbb E|X_i|\leqslant \max_{1\leqslant j\leqslant N}\mathbb E|X_i|+1+\mathbb E|X|.$$


1

May I ask what is your student number? This is clearly for Introduction to Statistics 35151 at UTS, and it is an assignment that is meant to be your own individual work or from group members. The assignment is due at 11am in the morning, I hope you finish it according to academic policies outlined by UTS.


4

If $x$ and $y$ are independent Gaussian random variables, then the linear combination $$z(t)= x\cos(wt)-y\sin(wt)$$ is also Gaussian. Can you calculate the mean and variance of $z(t)$?


1

You need to know Jacobian and change of variables. I will do the quotient, product is almost identical. Let us define the following transformation $Z=X/Y$ and $T=Y$. We are going to find joint density $p(Z,T)$ by change of variables from $X,Y$ to $Z,T$. So inverse transformation is, $$ X=YT\\ Y=T $$ and Jacobian of this transformation is given by, $$ ...



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