Tag Info

Hot answers tagged

3

You should have something like: $$ E(|X-Y|^a)=\iint_{x>y}(x-y)^a\;dxdy+\iint_{y>x}(y-x)^a\;dxdy=2\iint_{x>y} (x-y)^a\;dxdy $$ Now: $$ \iint_{x>y}(x-y)^a\;dxdy=\int_{y=0}^1\int_{x=y}^1(x-y)^a\;dxdy=\frac{1}{(a+1)(a+2)} $$ So: $$ E(|X-Y|^a)=2\iint_{x>y}(x-y)^a\;dxdy=\frac{2}{(a+1)(a+2)} $$


2

There is no conditional density, since with probability $1/n$, $Y_1 = z$. With probability $1-1/n$, $Y_1$ is uniform on $[0,z]$. So the conditional law is a mixture of an atom at $z$ and a continuous distribution supported uniformly on $[0,z]$.


2

We have $\displaystyle Y=X_1-X_2+1\sim N(0,3)$ (assuming independence). Therefore $\displaystyle \sqrt{c} Y\sim N(0,1)$ if we choose $\displaystyle c=\frac{1}{3}$, that is $\displaystyle \frac{1}{3} (X_1-X_2+1)^2 \sim \chi_{(1)}^2$.


1

We have $64$ cells: $36$ cells have $4$ neighbors each $24$ cells have $3$ neighbors each $ 4$ cells have $2$ neighbors each The total number of ways to choose $2$ out of $64$ cells is $\binom{64}{2}=2016$. The number of ways to choose $2$ adjacent cells is $\frac{36\cdot4+24\cdot3+4\cdot2}{2!}=112$. So the probability of choosing $2$ adjacent cells is ...


1

Another way to count adjacent cells is to count internal edges (borders between to cells). For each internal edge, there is a pair of adjacent cells. Horizontally there are $7$ lines of internal edges, each with $8$ internal edges; for a total of $56$ horizontal internal edges. Likewise there are $56$ vertical internal edges. Thus there are $112$ ...


1

Writing $$\left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k = \left( \frac{1}{n} \sum_{j=1}^n X_j \right) \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1}$$ we get $$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \frac{1}{n} \sum_{j=1}^n \mathbb{E} \left[ X_j \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right].$$ Since the ...


1

$E[XY]$ is not necessarily $E[X]E[Y]$ if $X$ and $Y$ are not independent. Here the number of tails is dependent on on the number of heads. In this case, $n^2/4$ is an upper bound on the product random variable (consider maximising $h(n-h)$ by varying $h$), and the maximum possible value if $n$ is even. $0$ is the minimum possible, and has positive ...


1

$\binom{12}{1}$ is the choice of the fifth card's rank; it can be any rank but the one chosen for four of a kind, so 12 remain. $\binom{4}{1}$ is the choice of the fifth card's suit. It can be any of the four. Usually I see the four of a kind lineup a little more succinctly: $$\binom{13}{1}\binom{48}{1}$$ This is equivalent, because $\binom{4}{4}=1$, ...


1

I would say there are $13\cdot 48$ such hands -- first you choose which of the 13 kinds it is you have all four of; then you choose one of the remaining 48 cards to bring the hand up to five cards. $\binom{12}{1}\binom{4}{1}$ seems to be a roundabout way of counting the 48 remaining cards. But I don't see where $\binom{13}{4}$ in the text you quote comes ...


1

$I(F) := \int_1^\infty \frac{F(y)}{y^2} dy$ $F$ is nonnegative and we know that $\int_1^\infty \frac 1 {y^2} dy = 1 $. So if $F(x) \leq 1 \forall x\geq 1$ then $$I(F) \leq \int_1^{\infty} \frac{dy}{y^2} = 1.$$


1

Starting from the probability $$ \prod_{j=0}^n\left[1-\frac{j}{n^3}\right] $$ and repeatedly using the relation that for $a,b>0$ $$ (1-a)(1-b) < 1-a-b $$ we get a lower bound of $$ 1 - \sum_{j=0}^n \frac{j}{n^3} = 1 - \frac{1}{n^3}\sum_{j=0}^n j = 1 - \frac{n+1}{2n^2} $$ And of course there is an upper bound of $1$.


1

You are seeking the CDF of the unconditional first machine repair time? If so, your solution is correct if only one machine can be repaired at the same time. Otherwise there are possible events of the form "machine A breaks down first and starts repairing, but machine B breaks down and is repaired before the repairing of machine A has been completed". ...


1

We flip $2k$ heads ($0\leq 2k\leq n$) with probability: $$\binom{n}{2k}\,p^{2k}(1-p)^{n-2k} $$ hence the probability of an even number of heads is given by: $$\begin{eqnarray*} \sum_{k=0}^{\lfloor ...


1

Your final answer is correct, to show another way: $$P(\text{of not drawing an acer})=\left ( \frac{6}{10} \right )\left ( \frac{5}{9} \right )\left ( \frac{4}{8} \right )\left ( \frac{3}{7} \right )\left ( \frac{2}{6} \right )=\frac{1}{42}$$ $$P(\text{of drawing at least one acer})=1-\frac{1}{42}=\frac{41}{42}$$


1

Here is a way to get the answer using more basic principles. $$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$ This much you already know. Then $$\begin{align*} F_U(u) &= \Pr[U \le u] \\ &= \Pr[Z^2 \le u] \\ &= \Pr[-\sqrt{u} \le Z \le \sqrt{u}] \\ &= \Pr[Z \le \sqrt{u}] - \Pr[Z \le -\sqrt{u}] \\ &= ...



Only top voted, non community-wiki answers of a minimum length are eligible