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4

I will mark the set of grades by the three people as $\{a, b, c\}$, (where order doesn't matter). Note that, if $x, y, z$ are distinct, then: \begin{align} \Pr[\{x, x, x\}] &= 1 / 5^3 \\ \Pr[\{x, x, y\}] &= 3 / 5^3 \\ \Pr[\{x, y, z\}] &= 6 / 5^3 \end{align} Final Grade = 1 Requires $\{1, 1, 1\}$ or $\{1, 1, a\}$ for some $a \geq 2$. Ie, ...


2

Imagine taking all the balls out in some random order and lining them up in the order they appear. The $r$ red balls act as separators for $r+1$ 'runs' of white balls (where a 'run' may be empty, contrary to the usual meaning of a run). Thus $w$ white balls are distributed among these $r+1$ runs. So the average run of white balls is of length ...


2

The gamma distribution is usually determined by two parameters, namely the shape parameter $k$ and the scale parameter $\theta$ (following wikipedia). Let me outline how I would do it. What you need is to translate your information about the mean and variance into information about the parameters $(k,\theta)$. But this can be done, since both the mean and ...


2

Is it good answer? $$X=\{0,1,2\}, Y=\{0,1\}$$ $$P(Y=1)=1/2, P(Y=0)=1/2, P(X=0)=1/4, P(X=1)=1/2, P(X=2)=1/4$$ $$P(X+Y=0)=1/8, P(X+Y=1)=3/8, P(X+Y=2)=3/8, P(X+Y=3)=1/8$$ $$P(X-Y=-1)=1/8, P(X-Y=0)=3/8, P(X-Y=1)=3/8, P(X-Y=2)=1/8$$ $P(XY=0)=5/8, P(XY=1)=2/8, P(XY=2)=1/8$


2

Use a generating function. For one generation (no pun intended), the generating function is $$ F(z) = \frac{z+z^2+z^3}{3} $$ since there is a one-third probability of producing either one ($z$), two ($z^2$), or three ($z^3$) offspring. To determine the generating function for the grandchildren, we simply write $$ \begin{align} F(F(z)) & = ...


1

The cumulative distribution function is never convex on all of $\mathbb R$. Do you mean it is convex on some interval? Have you tried looking for examples?


1

First, you should be able to show that combined production has 10/300 = 3.33% defectives. Then, the number of defectives in a sample of size $n = 10$ is $X \sim Binom(10, 10/300),$ so you can use the binomial formula to find $P(X = 2).$ According to my calculations your answer should be a little smaller than 0.04.


1

You want to find $c$ such that $$-1\le \dfrac{\sqrt{(Y-2/c)^2-4}}{2} \le 1$$ (and in particular is real) for $0 \le Y \le 1$. Equivalently, $$ 4 \le (Y - 2/c)^2 \le 8$$ for $0 \le Y \le 1$. So either $2 \le Y - 2/c \le 2 \sqrt{2}$ in that interval, or $-2\sqrt{2} \le Y - 2/c \le -2$. Now unfortunately, $2\sqrt{2} - 2 < 1$, so this can't ever be true ...


1

One could say it's "differentiable at $0$ within the domain $[0,\infty)$". Within that space, $0$ can be approached only from the right, so "$\lim\limits_{x\to0}$" could only mean it's approaching $0$ from the right. Without something like that interpretation, the information you have been given is, as you have noted, contradictory. That is probably what ...


1

What you're looking for is the Rayleigh distribution (distribution of the norm of two centered and independent gaussian RVs) : http://en.wikipedia.org/wiki/Rayleigh_distribution You might also want to look up the $\chi^2$ distribution (distribution of the squared norm) : http://en.wikipedia.org/wiki/Chi-squared_distribution


1

Let $R$ denote the distance of the point $(X,Y)$ from the origin. Draw a circle of radius $a$ centered at the origin. Then, $$P\{R \leq a\} = F_R(a) = \iint_{x^2+y^2\leq a}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy.$$ Convert from rectangular coordinates $(x,y)$ to polar coordinates $(r,\theta)$ hopefully not forgetting the mantra $r\,\mathrm dr\,\mathrm d\theta$ ...


1

If $0 < Y < 1$, then $0^2 < Y^2 < 1^2$, so it follows that $0 < Z < 1$. More generally, if $a < Y < b$, then we must address three cases: $$\begin{align} 0 \le a < b, \\ a < 0 < b, \\ a < b \le 0. \end{align}$$ In the first case, it should be clear that $$a^2 < Z < b^2.$$ In the second, because $0 \in (a,b)$, we ...


1

The distance is chi-square distributed. Intuitively, look at the distance between the two normal random variables as an error (for example, distance between randomly sampled point and the mean). We know that this error is distributed chi square. Here's a more rigorous description: ...


1

You have the limits of integration wrong. You should have: $$ F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{x} f(u,v)\; du\; dv $$ But your density is zero when $u\ge v$, so your integral becomes: $$ F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{\mbox{min}(x,v)} e^{-v}\; du\; dv $$ So: $$ F_{X,Y}(\infty,\infty)=\int_{v=0}^{\infty}\int_{u=0}^{v} e^{-v}\; du\; ...


1

Good attempt, but you have to consider your support (i.e., the region of integration) for this question. The shaded region is the region in which $0 < x < y$. Now consider the following example: The shaded region is $$\mathbb{P}\left(X \leq 2, Y \leq 4\right) = \int_{0}^{2}\int_{x}^{4}f_{X,Y}(x,y)\text{ d}y\text{ d}x\text{.}$$ You can generalize ...


1

We are given the piecewise cumulative distribution function $$\Pr[X \le x] = F_X(x) = \begin{cases} 0, & x < 0 \\ 1/3, & 0 \le x < 1 \\ 2/3, & 1 \le x < 2 \\ 1, & 2 \le x. \end{cases}$$ What the first two cases tell us is that $\Pr[X = 0] = 1/3$, since $$\Pr[X \le 0]$$ corresponds to the choice $x = 0$ in the above function, thus we ...


1

A breaks 5 times as often as B. So for every 6 broken devices, 5 times it will be A, and 1 time it will be B. The broken device was created by A with a $\frac{5}{6}$ probability. Multiplying $\frac{5}{6}$ by the chance A will break the next device, $\frac{1}{20}$, gives $\frac{1}{24}$. Now the broken device has a 1/6 probability of being created by B. So ...


1

In your confidence interval, you are using $\bar x = 7$, which is inconsistent with your earlier notation. It should be $\bar d = \bar y - \bar x= 7.$ Also, to be clear, 16 is the variance for the difference in scores for one individual, so the variance for $\bar d$ is $16/3$ and the standard deviation used in the confidence interval is $4/\sqrt{3},$ as you ...


1

I have a simpler proof. I hope this is ok. Let $Y = X - a$ be a random variable. Now note that due to symmetricity $Y$ and $-Y$ have the same distribution. That implies $$E[Y^3] = E[(-Y)^3]$$ This implies $E[Y^3] = 0$.


1

Solve $a$ and $b$ from the equations: $ra+sb=1$ $a+b=c$ Here $r$ and $s$ are allready known to you and $c$ denotes a fixed number. This comes to finding (linear) expressions for $a$ and $b$ in $c$. Then check for wich $c$ the solution $\langle a,b\rangle$ satisfies $a\geq0\wedge b\geq0$. For each of them the corresponding $f(t)$ is a PDF.


1

Hint: Here is a visual of the region of integration you seek: The entire colored region is the support used for integrating the density functions since $x, y \in (0, 1)$. Now recall that $$\mathbb{P}\left(Y \geq X \mid Y \geq \dfrac{1}{2}\right) = \dfrac{\mathbb{P}\left(Y \geq X \text{ and } Y \geq \dfrac{1}{2}\right)}{\mathbb{P}\left(Y \geq ...


1

We need to take a weighted average of events. That is, $\frac{r}{r+w}$ percent of the time, we will only have to draw one ball. Then $\frac{r^2}{(r+w-1)(r+w)}$ percent of the time it takes 2. This goes on and on and we need the weighted average, which, in closed form is$$ ...


1

The probability that any one white ball gets picked is $1/(r+1)$. The average number of white balls that get picked is $w/(r+1)$.


1

Write out the summation terms, then perform the division. With the quotient, differentiate term wise and conclude that f'(y) is negative on the given interval and variable bounds.



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