# Tag Info

1

To find the p.d.f of the ratio $\frac{Y}{X+Y}$, let us first write its c.d.f. Since $X$ and $Y$ are always positive, their ratio is also positive and, therefore, for $0\leq t\lt1$ we can write: $P\left(\frac{Y}{X+Y}\leq t\right)=P\left(Y\leq \frac{t}{1-t}X\right)=\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx$ as ...

0

You know that $K$ of the $2N$ computers were checked until the $N$-th of one version was found.   Thus the $K$-th computer is of that type, and $0\leq K{-}N < N \leq K < 2N$. Now the probability of selecting $N{-}1$ of $N$ computers of one version, and $k-N$ of the $N$ computers of the other, in any order, then selecting the $1$ remaining ...

1

For any positive integer $n$, we have $T_n-S_n = g(T_n,S_n)$ where $g:\mathbb R^2\to \mathbb R$ is the map $(x,y)\mapsto x-y$. Given $t\in\mathbb R$, it is clear that $$g^{-1}((-\infty,t]) = \{(x,y):x-y\leqslant t\}$$ is a Lebesgue-measurable set in $\mathbb R^2$, and so $g$ is a measurable function. Since $\sigma(g(T_n,S_n))\subset \sigma(T_n,S_n)$, ...

0

Let $\mu=E(X)$. Then $cov(X)=E(XX^T)-\mu\mu^T$ and $cov(AX)=E((AX)(AX)^T)-(A\mu)(A\mu)^T=A(E(XX^T)-\mu\mu^T)A^T=Acov(X)A^T$.

2

For $x>1$, we have $$\mathbb P(\xi_n\leqslant x) = \mathbb P\left(\bigcap_{i=1}^n \left\{\eta_i\leqslant x\right\}\right)=\prod_{i=1}^n\mathbb P\left(\eta_i\leqslant x \right) = \left(1 - x^{-\alpha}\right)^n.$$ Hence \begin{align} \mathbb P(\zeta_n\leqslant x) &= \mathbb P\left(\xi_n n^{-\frac1\alpha}\leqslant x\right)\\ &= \mathbb ...

0

We first do it more or less in the way you attempted. There are $\binom{10}{7}$ equally likely ways to choose $7$ wines. First we find the number of ways to choose $0$ Pinks. There is only one way, so $\Pr(Y=0)=\frac{1}{\binom{10}{7}}$. For the future, note that this is $\frac{\binom{3}{0}\binom{7}{7}}{\binom{10}{7}}$. So now we know $\Pr(Y=0)$. Next we ...

0

Close, but a few details need attention. Assuming the selection is unbiased and without repetition, then the probability of choosing $y$ of $3$ white wines, and $7-y$ of $7$ not white wines, out of all the ways to choose $7$ of $10$ wines is: $$\Pr(Y=y) \;=\; \mathbf 1_{y\in\{0,1,2,3\}}\cdot{\dbinom{3}{y}\dbinom{7}{7-y}}\Big/{\dbinom{10}{7}}$$ So the ...

1

Notice that because of independence: $$P\left(X^2 < \frac{1}{2}, |Y| < \frac{1}{2} \right) = P\left(X^2 < \frac{1}{2}\right) P\left(|Y| < \frac{1}{2} \right)$$ Analyzing the $X$ term: P\left(X^2 < \frac{1}{2}\right) = P \left( -\frac{1}{\sqrt{2}} < X < \frac{1}{\sqrt{2}} \right) = ...

0

If $Z\sim\text{unif}(-1,1),$then $$f_{|Z|}(z) = 1.$$ You can take this to mean that $|Z|\sim\text{unif}(0,1)$. In your case $$P(X^2<1/2,|Y|<1/2) = P(|X|<1/\sqrt 2)\cdot P(|Y|<1/2) = (1/\sqrt 2)(1/2),$$ since $X$ and $Y$ are independent.

2

Since $X$ and $Y$ are independent, we have $$\Pr[(X^2 \le 1/2) \cap (|Y| \le 1/2)] = \Pr[X^2 \le 1/2]\Pr[Y \le 1/2].$$ Since they are uniform on $[-1,1]$, we then have $$\Pr[X^2 \le 1/2] = \Pr[-1/\sqrt{2} \le X \le 1/\sqrt{2}] = \frac{1/\sqrt{2} - (-1/\sqrt{2})}{1 - (-1)} = \frac{1}{\sqrt{2}},$$ and $$\Pr[|Y| \le 1/2] = \Pr[-1/2 \le Y \le 1/2] = ... 2 Hint: X^2< \frac12\iff |X|<\frac1{\sqrt2}. If X and Y are independent, then |X| and |Y| are also independent. Complete the sentence: if A and B are independent, then P(A, B) = ... 0 You're wrong with your initial assumption. Actually, we have$$P (X_n = k) = \frac {6 - |k - 7|} {36}.$$Also, the probability you ask for is the sum$$\sum_{k = 1}^{12} (P(X_n = k)^2) = \sum_{k = 1}^{12} \left ( \frac {6 - |k - 7|} {36} \right )^2 = \frac {1} {1296} \sum_{k = 1}^{12} (36 + (k - 7)^2 - 12 |k - 7|) = \frac {12 \cdot 36 + 146 - 12 \cdot 36} ...

0

Hints: $\mathbb EX=P(X\geq1)+\dots+P(X\geq N)$ $P(X\geq k+1)+P(X\leq k)=1$ edit: $$\mathbb{E}X=\sum_{i=1}^{N}ip_{i}=\sum_{i=1}^{N}\sum_{k=1}^{i}p_{i}=\sum_{k=1}^{N}\sum_{i=k}^{N}p_{i}=\sum_{k=1}^{N}P\left(X\geq k\right)=$$$$\sum_{k=1}^{N}\left(1-P\left(X\leq k-1\right)\right)=N-\sum_{k=1}^{N}P\left(X\leq k-1\right)=N-\sum_{k=n}^{N-1}P\left(X\leq ... 1 The answer to your first question is yes: if c \ne 0 and c \in \mathbb R, then cX \sim \operatorname{Normal}(c\mu, |c|\sigma) if X is normal with mean \mu and standard deviation \sigma. This is because the normal distribution belongs to a location-scale family: its PDF is$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad ...

0

The answer is negative. If $(r_n)$ is lacunary sequences (that is $r_{n+1}>ar_n$ for some $a>1$), then for any probability preserving transformation $T\colon X\to X$ and any $\delta>0$ there is $A\subset X$ with $0<\mu(A)<\delta$ such that $$\limsup_n \frac{1}{n} \sum_{k=0}^{n-1} \chi_A (T^{r_k}x)=1 \quad \text{ for a.e. x\in X.}$$ Akcoglu ...

1

The probability of extinction is the smallest positive root of $$G_O(z)=z$$ Where $O$ denotes the offspring distribution, and $G_O(z)$ its generating function at $z$. It is easily seen that $G_O(0)$ is the probability of extinction in the first generation. Second, if you know about generating functions, then you know that the sum: , where $X$ is ...

2

For any $p \geq 1$, we have $$|x+y|^p \leq 2^p (|x|^p+|y|^p),$$ and therefore \begin{align*} \mathbb{E}(|X_n-X|^p \mid \mathcal{F}) &\leq 2^p \mathbb{E}(|X_n|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}) \\ &\leq 2^p \mathbb{E}(|X|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}). \end{align*} This shows that ...

0

Just to be clear: The expectation of an Indicator Random Variable is the probability of it being 1. \begin{align}\mathsf E(X_i) & = 0\cdot \mathsf P(X_i=0)+1\cdot \mathsf P(X_i=1)\\ & = \mathsf P(X_i=1)\end{align} $X_i$ is the indicator that red ball #$i$, for $i\in\{1..10\}$, is one of the $12$ out of $30$ balls drawn. Imagine we lay the ...

1

As $0 \leq a, \bar{a} \leq 1$, we have $$(a-\bar{a})(a + \bar{a}) \leq |(a-\bar{a})(a+\bar{a})| = |a-\bar{a}| \cdot \underbrace{|a+\bar{a}|}_{= a+\bar{a} \leq 2}.$$ Taking expectation on both sides, proves the inequality.

1

$$E[(a-\bar{a})(a+\bar{a})]\leq E[|a-\bar{a}|(a+\bar{a})]\leq 2E[|a-\bar{a}|],$$ where the second inequality follows from the fact that $(a+\bar{a})\leq 2$.

1

No, there is nothing to be said about the Cesàro averages of the whole sequence. They can be as bad as the sequence itself. Indeed, given any weakly convergent sequence $\{f_n\}$, we can consider another sequence $\{g_n\}$ defined as $$f_1,f_1, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_2,f_2,f_2,f_2, f_3, \dots$$ where the term $f_n$ appears ...

-1

Important inequalities (Probability w/ Martingales): 1, 2 $$\liminf x_n > z \to \liminf(x_n > z)$$ $$\liminf x_n < z \to \limsup(x_n < z)$$ 'if' Suppose $\forall \epsilon>0$, $$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$$ By BCL1, we have $$\textrm{P}\left(\limsup ... 0 For the employer to know the order in which the computers were checked, he must have been noting the first computer as version A and the first computer with a different version than A as B and then forgetting what version A and B correspond to. An example of a sample path of him noting: A, A, B, A, B, B, A, ... When he has N of the same system ... 1 Suppose that we have some points y_{1},y_{2},\dots,y_{n} and we have the model$$y_{i}=x_{i}^{T}\beta+\xi_{i}where the \xi_{i} are i.i.d. \mathcal{N}(0,\sigma^{2}) noise terms, \beta\in\mathbb{R}^{p} and x_{i}\in\mathbb{R}^{p}. This can be rewritten as y=X^{T}\beta+\xi where the columns of X are x_{1},\dots,x_{n} and \xi is a ... 0 Note that we have \begin{align*} \def\E{\mathbf E}\E[y^* C_{yy}^{-1}y] &= \E\left[\sum_{i,j} \bar y_i \bigl(C_{yy}^{-1}\bigr)_{ij} y_j\right]\\ &= \sum_{i,j} \bigl(C_{yy}^{-1}\bigr)_{ij} \E[\bar y_i y_j]\\ &= \sum_{i,j} \bigl(C_{yy}^{-1}\bigr)_{ij} \bigl(C_{yy}\bigr)_{ji}\\ &= \operatorname{tr} C_{yy}^{-1}C_{yy}\\ &= n. ... 1 It is a standard result in measure theory that for nonnegative functions f\int_A f d \mu = \lim_{n \to \infty} n \mu(f^{-1}([n,\infty)) + \sum_{k=1}^{n^2-1} \frac{k}{n} \mu(f^{-1}([k/n,(k+1)/n)).$$(The details of the partitioning are not so important; the important matter is that the mesh size goes to zero and the upper bound goes to infinity.) For ... 1 \int_{-\infty}^\infty\int_{-\infty}^yf(x,y)\;\;dx\;dy 0 Yes, we will have to divide into cases. There are two uninteresting cases, (i) y\le 0 and (ii) y\ge 2. If y\le 0, then F_Y(y)=\Pr(Y\le y)=0. If y\ge 2 then \Pr(Y\le y)=1. The other two cases are (iii) 0\le y\le 1 and (iv) 1\lt y\lt 2. Case (iii): We have Y\le y if and only if -y\le X\le y. The density function of X on the interval ... 0 You were on the right track. The only problem with your solution is that you don't take in consideration that P(Y<1∣X<1)=1 since 0<y<x. If you take that into consideration, you will find:$$\ P(Y\lt 1) = \int_{0}^{3} P(Y \lt1 \mid X = x) \ f_x(x) \ dx\ P(Y\lt 1) = \int_{0}^{1} 1* \ f_x(x) \ dx + \int_{1}^{3} P(Y \lt1 \mid X = x) \ ...

0

I believe the abuse of notation which is causing the confusion involves the notion of the preimage of an element of the codomain of the random variable. https://en.wikipedia.org/wiki/Image_(mathematics) Consider fact 2.13 without the abuse of notation. Let $X$ be a random variable on $S$. Then E(X) = \sum\limits_{x \in \mathbb{R}}x ...

0

If $g$ denotes the conditional PDF then it is prescribed by $x\mapsto\frac{32}{15}x$ if $\frac14<x<1$ and $x\mapsto0$ otherwise. $$P(X>\frac34|Y=\frac14)=\int\chi_{(\frac34,\infty)}g(x)dx=\int_{\frac34}^1\frac{32}{15}xdx$$

1

When you calculate a function (here $f_{X\mid Y=1/4}$) try always writting also its domain. Here the domain of $f_{X\mid Y=1/4}$ is: $$f_{X\mid Y=1/4}(x \mid 1/4)=\begin{cases}\frac{32}{15}x, & 1/4<x<1 \\0, & \text{else } \end{cases}$$ Now you see that $$P(X >3/4 \mid Y=1/4)=\int_{3/4}^{1}\frac{32}{15}x\ dx$$

2

We have $$E(X^2)=\int_0^\infty x^2 \lambda e^{-\lambda x}dx=\frac{2}{\lambda^2}$$ So, $$Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}$$

1

No, there's no need to calculate the density of $V$.   Don't make unnecessary work for yourself. You have $X \sim \mathcal U\{0,1,2,3,4,5\}$, and $R\sim\mathcal U(0.04;0.08)$ and that $X$ and $R$ are independent. You want to calculate $\mathsf E(X \,\mathsf e^{2R})$.   That is: $$\mathsf E(X\, \mathsf e^{2R}) = \sum_{x=0}^5 ... 1 One step at a time. First step. If 0 \lt X^2 then \frac 1{X^2}\lt \infty . Because the inverse of every positive number is a finite number. Second step. If 0 < X^2\leq 1 then 1\leq \frac{1}{X^2}. Because the inverse of every positive number no greater than one must be a positive number no lesser than one. Put it ... 0 Hint: If X=aY+b then how are X^* and Y^* related? 0 I agree that the assertion looks a little odd, but it is correct. Using your analysis, we have that$$\Pr(X\le k\le \alpha X)=\Pr\left(X\ge \frac{k}{\alpha}\right)-\Pr(k\le X).$$(Note the correction in the second term.) The first term on the right is then 1-\Pr(X\le \frac{k}{\alpha}) and the second term is 1-\Pr(k\ge x). Subtract and note the ... 1 Outline: Let a be the probability that a randomly chosen female Smurf is between 1 and 1.3, and let b be the corresponding probability for male Smurfs (Smurves?). Then our required probability is (0.6)a+(0.4)b. 0$$P\{ Y \le y\} = P\left\{\frac{1}{Z}\le y\right\} = P\left\{Z \ge \frac{1}{y}\right\} = \int_{1/y}^\infty f_Z(z)\mathrm{d}z\implies f_Y(y) = \frac{\partial}{\partial y} P\{Y \le y\} = \frac{1}{y^2} f_Z(1/y)$$Alternatively, recall that densities of transformed random variables are in the ratio of the Jacobian of the transformation.$$z = ...

0

The change of variables formula is (from the chain rule): \begin{align} f_{g(Z)}(y) & = f_Z(g^{-1}(y)) \cdot \lvert \mathcal D_y\; g^{-1}(y)\rvert & \textrm{where g(z) is an invertable function} \\[2ex] f_Y(y) & = f_Z(1/y)\cdot\left\lvert\dfrac{\operatorname d y^{-1}}{\operatorname d y\quad}\right\rvert & \textrm{since g(z)=1/z} ... 1 By the properties of a continuous density, \mathbb P(0 < Y \le b) = \int_0^b f_Y(y) \, dy. $$Therefore, for Y = 1/Z,$$ \mathbb P(0 < Y \le y) = \mathbb P(Z \ge 1/y) = \int_{1/y}^{\infty} f_Z(z) \, dz = 1/2 - \int_0^{1/y} f_Z(z) \, dz. $$Differentiating with respect to y to recover Y's density,$$ f_Y(y) = \frac d{dy} \left( 1/2 - ...

0

Hint: If $B$ is an event in $F$ and $g:\Bbb R\to\Bbb R$ is bounded and continuous, then $\Bbb E[1_Bg(X)]=\lim_{n\to\infty}\Bbb E[1_Bg(X_n)]$. [For this you need to know that convergence in probability is preserved by composition with continuous functions.]

0

HINT: I'll go ahead and give you a hint on part A since you haven't shown much effort or thought in searching for an answer. What's your expected location? The average distance you would have to walk to either bus stop should be total possible distance minus your expected distance.

0

For your first question, we have $$P\left(X^2\leq y\right)$$ $$=P\left(\sqrt{X^2}\leq \sqrt y\right)$$ $$=P\left(\left|X\right|\leq \sqrt y\right)$$ $$=P\left(−\sqrt{y}\leq X\leq\sqrt{y}\right)$$ For your second question, we have $$-1\lt x\lt 1$$ Which can be rewritten as $$\left|x\right|\lt 1$$ Or $$0\leq\left|x\right|\lt 1$$ Have a look at absolute value.

1

$Z$ is nonnegative so $f_{Z}\left(x\right)=0$ if $x<0$. For $x\geq0$ we have: $$F_{Z}\left(x\right)=F_{X}\left(\sqrt{x}\right)-F_{X}\left(-\sqrt{x}\right)$$ Consequently: ...

0

While Ákos Somogyi's answer is correct, this is not the best approach to solve this kind of problem. There is a transformation of density formula, which says that if $X$ has density $f_X$ and $h$ is piecewise continuously differentiable and piecewise strictly monotone, then $Y$ has the density $$f_Y(y) = \sum_{x: h(x) = y} \frac{f_X(x)}{|h'(x)|}. \tag{1} ... 1$$ \mathbb{P}(X\in[-\sqrt{z},\sqrt{z}])=\mathbb{P}(X<\sqrt{z})-\mathbb{P}(X>-\sqrt{z})=\boxplus $$Now since X in uniform on [-1,2], the CDF of it is:$$ F_X(x)=\chi_{x\in(2,\infty)}+\frac{1}{3}(x+1)\chi_{x\in[-1,2]} $$THus by taking the range into consideration, we obtain:$$ ...

0

Since you said you've enumerated the outcomes for $X$, do the same for $Z$. Below I made a table for the values of both $X$ and $Z$. Can you now make the corresponding table for $W = XZ$?  \begin{array}{|c|c|c|c|c|c|c|} \hline X& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 4 ...

0

For each outcome of the two rolls, find the values of $X$ and $Z$ and the value of $W$ is given by their product.

-1

Let's say a coin is tossed once: we get the Expectation of number of heads as E[R] = no.of heads*prob(no.of heads) = 1*1/2 = 1/2. Now let n=2, E[R2] = 2*1/4 + 1*2/4 + 0*1/4 = 1. Now let n=3, E[R3] = 3*1/8 + 2*3/8 + 1*3/8 = 12/8 = 3/2. As we keep continuing this we observe that E[Rn] = n*1/2. This is the linearity of Expectation. That is, E[R] = ...

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