New answers tagged

1

I agree with both your calculations and get $\frac{1892}{23205}$ for both of them.


0

Scenario $B$ is equivalent to a scenario where each die has success probability $\frac23$ on the first roll and $\frac16$ on all further rolls, and is rerolled as long as it succeeds. We can use this to express the distribution as a single convolution, instead of the $(x+y)$-fold convolution that you wrote. The probability for $k$ of $m=x+y$ dice to succeed ...


2

There's a subtle flaw in your reasoning. You're effectively conditioning on which non-common gem type comes first. Your calculation for phase $2$ is correct: The non-common gem type already seen in phase $1$ has become irrelevant, and only the relative probabilities of the other two gem types matter. But this doesn't work in phase $1$, because the length of ...


1

Let us fix $\varepsilon\gt 0$. We want to find some $a$ such that $$\sup_n \mathbb P\left\{\left|X_n\right|\gt a\right\}\leqslant 2\varepsilon.$$ We assume that $\varepsilon$ is smaller than $1/2$. By assumption, we can find $a$ such that for each $n$, $$\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]\lt \varepsilon\mathbb ...


0

$\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y) = P(Y>X)$ $(\sum_\limits{x=0}^{99} P(x)\sum_\limits{y=x+1}^{100} P(y))y = E[y|y>x]P(y>x) = \frac 12 E[Y]$ I am having a hard time remembering exactly why this is....more later... $\frac {E[Y]}{2P(Y>X)} = \frac {E[Y]}{1-P(X=Y)} = \frac{50}{1-0.0563} = 52.98$ If we relax the condition ...


1

I think the first is correct (law of total variance), for the second I suggest law of total probability $P(Y) =E[P(Y\mid X)]$: $$P(Y\leq 1) = E[P(Y \leq 1\mid X)] = \int_{X} P(Y \leq 1 \mid x)\cdot f(x) dx$$ $Y\mid X = U \sim N(x,1). \ $, We know that $U <c, U \sim N(\mu,\sigma^2) \Leftrightarrow Z<\frac{c-\mu}{\sigma}$, where $Z \sim N(0,1)$. So: ...


0

EDIT: I made the mistake of including the case where $|X_1-X_2|=0$ in $E(|X_1-X_2|)$, when that has no impact on it. I've changed that now. I have a closed form expression that is somewhat inelegant, if anyone is interested. The distribution is clear but I'll proceed as if we didn't know what it was.$$$$ The space of events is $\{0,1\}^{200}$. If $X_1$ is ...


4

To approach the problem via normal approximations, let's first think about normal distributions in general to understand the distribution of the maximum we assume that $X_1,X_2$ are independently distributed as normal variables with mean $\mu$ and standard deviation $\sigma$. We see that $$P(\max (X_1,X_2)<\mu+t\sigma)=P(X_1<\mu+t\sigma)\times ...


5

Let $X$ be the random number of heads flipped by Player $1$ and $Y$ be the random number of heads flipped by Player $2$. Suppose $X$ and $Y$ are IID binomial random variables with common parameters $n$ and $p$. Then the desired expectation is $$\operatorname{E}[\max(X,Y)] = \sum_{x=0}^n \sum_{y=0}^n \max(x,y) \binom{n}{x} p^x (1-p)^{n-x} \binom{n}{y} p^y ...


1

A sanity check: verify you get the right expectation. (This is not sufficient, but at least it's a necessary condition for correctness). Since $Y\sim U(X,1)$, we have $$\mathbb{E}[Y\mid X] = \frac{1}{2}(1+X)$$ and therefore $$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\mid X]] = \frac{1}{2}(1+\mathbb{E}[X]) = \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$$ since ...


1

Why do we use $G(x)=1−F(x)$ instead of just $F(x)$? Because the memoryless property is that: $\mathsf P(X>s+t\mid X>s)=\mathsf P(S>s)$ So we can use this to state: $$\begin{align}\mathsf P(X>s+t) =&~\mathsf P(X>s)\mathsf P(X>s+t\mid X>s)\\=&~\mathsf P(X>s)\mathsf P(X>t)\\[2ex] 1-F_X(s+t)=&~ ...


0

In the proposed order: You can use $G(x)$ instead of $(1-F(x))$ just to make you demonstration less clumsy by not carrying that difference around. For instance, the first line would be more readable: Try $s=t$, this gives us $G(2t)=G(t)^2$, $G(3t)=G(t)^3$, ..., $G(kt)=G(t)^k$ Try $s=t$, this gives us $1-F(2t)=(1-F(t))^2$, $1-F(3t)=(1-F(t))^3$, ..., ...


1

The probability of selecting $y_1$ of $N_1$ white balls in a sample of $n$ from $N$ (where $3\leq n< N_1$ is: $$\mathsf P(Y_1=y_1) = \dfrac{\binom{N_1}{y_1}\binom{N-N_1}{n-y_1}}{\binom{N}{n}}$$ Which is a Hypergeometric Distribution, whose mean and variance you should know; or you can obtain by using indicators. Let $Y_{1,i}$ be the indicator that the ...


0

(Essentially copied from here) $$2 d_H(P,Q)^2 = \sum_i (\sqrt{p_i} - \sqrt{q_i})^2= \sum_i p_i+q_i - 2 \sqrt{p_i qi}=2(1- \sum_i \sqrt{p_i qi}) \tag{1}$$ But since $x-1 \ge \log x$ $$1- \sum_i \sqrt{p_i qi}\le - \log ( \sum_i \sqrt{p_i qi})= - \log E_p \sqrt {Q/P} \tag{2}$$ And by Jensen inequality $\log E(\sqrt {\circ})\ge E (\log \sqrt{\circ}) ...


1

Since this is a discrete distribution, $Y=0,1,2$ If $Y=0$, then it means two heads are next to each other. So, $P(Y=0|X=2)=\frac{n-1}{{n\choose 2}}=\frac{2}{n-2}$ (since there are n-2 ways to choose the two consecutive heads, and $n\choose 2$ ways to randomly choose 2 positions for 2 heads among n heads) $Y=1$ is not possible, because you cannot have one ...


0

You either have $0$ isolated heads or $2$ isolated heads. There are $\begin{pmatrix} n \\ 2 \end{pmatrix}$ ways to place exactly $2$ heads among the coin tosses. Of which $n-1$ of these choices gives you non-isolated heads.. Hence $$Pr(Y=0|X=2)=\frac{n-1}{\begin{pmatrix} n \\ 2 \end{pmatrix}}=\frac{2}{n}$$ $$Pr(Y=2|X=2)=1-Pr(Y=0|X=2)=\frac{n-2}{n}$$


0

We have $$C(t)=\frac{6}{10}+\frac{e^{it}}{10}+\frac{e^{-it}}{10}+\frac{e^{2it}}{10}+\frac{e^{-2it}}{10}.$$ If $X$ has characteristic function $C$, then the term $e^{it}/10$ comes from the fact that $\mathbb P\{X=1\}=1/10 $, and similarly for the other values. More generally, we can identify characteristic functions which are convex combinations of ...


1

(a) Yes, the sum of $n$ iid Geometric Distributed Random Variables has a Negative Binomial distribution, and that is the right moment generating function for the given one. (b) is okay, and see also (d) below. (c) Well, $\mathsf e^{4t/3}$ is the moment generating function for a Degenerate Distribution. In this case ...


0

Think of it this way: The parameter of the exponential distribution is the rate. $X$ is the time till the next Poisson point event occurring at rate $1$. $Y=X/\lambda$ is thus the (shorter for $\lambda>1$) time till the next Poisson point event occurring at (faster for $\lambda>1$) rate $\lambda$.


2

This is because both $X_n$ and $Y$ only take values from a set of two points $\{3, 8\}$, which means given any positive $\epsilon \in (0, 5]$, the set \begin{align} \{\omega: |X_n(\omega) - Y(\omega)| \geq \epsilon\} & = \{\omega: X_n(\omega) = 8, Y(\omega) = 3\} \cup \{\omega: X_n(\omega) = 3, Y(\omega) = 8\} \\ & = \{\omega: X_n(\omega) \neq ...


0

You could easily multiply. If $c>0$, then the result given implies $$cX\sim\text{Exp}(1/c).$$ It seems like matter of style to me. In general, we have that if $X\sim\text{Exp}(\lambda)$, and $\lambda, c>0$, then $$cX\sim\text{Exp}(\lambda/c).$$ Or as your book might put it, $$\frac{X}{c}\sim\text{Exp}(c\lambda).$$


2

Method 1: Using the Law of Total Probability. Let $S$ be the event of selecting a child with a sister. Let $F_{\rm GGG}, F_{\rm GGB}, F_{\rm GBB}, F_{\rm BBB}$ be the event of selecting a child from the relevant family structure; given that we know we are selecting from families with three children. The count of girls in a family has a Binomial ...


0

A card is given to Player $2$ exactly if it comes before all cards smaller or equal to it and before the $3$ of hearts. In a standard deck, there are $13$ different values and $4$ suits of each value, and a card with the $k$-th smallest value has $4k-1$ cards smaller or equal to it. By symmetry, the probability that the card comes first out of the ...


2

Both methods are fine. You didn't really specify the problem, but I gather you meant to imply that a child is selected from the family uniformly randomly, and that the genders of the three children are independent. In that case, both your arguments are correct.


0

Since $4500000$ doesn't divide $80000000$, the assumption that each value has approximately the same number of rows doesn't really specify how often the value under consideration occurs. Let's round to the nearest integer and assume that it occurs $18$ times (keeping in mind that there are other values that occur at most $17$ times). There are ...


0

If $A$ is a matrix with $\left(U,V\right)^{T}=A\left(X,Y\right)^{T}$ then $\mathbb{E}\left(U,V\right)^{T}=A\mathbb{E}\left(X,Y\right)^{T}$. If $\Sigma$ is the covariancematrix of random vector $\left(X,Y\right)^{T}$ then $A\Sigma A^{T}$ is the covariance matrix of random vector $\left(U,V\right)^{T}$. $\left(U,V\right)^{T}$ has normal distribution, so its ...


0

For example, $P(0,0)=\frac12\times\frac12\times\frac12=\frac18$ because it is calculating the probability that all three tosses are heads. $P(1,0)$ is a conditional probability given that the first toss is a head, which looks like $P(X=1|Y=0)=\frac{P(\text{One Tail})}{P(Y=0)}=\frac{\frac18}{\frac12}=\frac28$ Thus the joint distribution of (X,Y) can be ...


3

A lot of times, instead of working with the cumulative distribution functions, it's easier to work with moment generating functions. If you haven't seen these before, for a random variable $X$, the moment generating function is $M_X(t) = E\left(e^{tX}\right)$, if this expected value exists for $t$ in some neighborhood of $0$. One can take advantage of the ...


0

The methodology for such a process is as follows: 1) Generate i.i.d standard normals (X,Y,Z). You can think of this as a vector $z$ 2) Do a Cholesky Decomposition on your specified correlation matrix $R$. 3) Multiple your vector $z$ by your newly decomposed vector. This will generate 3 correlated standard normals. 4) These correlated standard normals can ...


3

As a typical example of what happens in a gap, look at $3/2$, which is in the first gap. The CDF value at $3/2$ is $$ \int_{-\infty}^{3/2}f_X(x)\,dx= \int_{-\infty}^00\,dx+\int_0^1\frac14\,dx+\int_1^{3/2}0\,dx= 0+\frac14+0=\frac14. $$ The rest of $F_X$ can be computed the same way (though the integral splits into more pieces when you go to later intervals).


0

$F_{X}(x) = \begin{cases} 0, & x \leq 0 \\ x/4, & \text{if 0 < x } \leq 1\\ 1/4, & \text{if 1< x }\leq 2\\ 1/4+(x-2)/4, & \text{if 2 < x } \leq 4\\ 3/4, & \text{if 4 < x } \leq 6\\ 3/4+(x-6)/4, & \text{if 6 < x }\leq 7\\ 1, & \text{elsewhere}\\ \end{cases}$ The CDF stays constant when the density is zero.


0

You want to answer the first question, "what is the probability of picking an orange?". Let's split this question into three mutually disjoint and exhaustive cases: What is the probability of picking an orange from the red box? The probability of picking the red box to begin with is $0.2$, and after that, the probability of picking an orange is ...


1

For the first question, you need to condition on whether the fruit is chosen from the red box, the green box, or the blue box and use the law of total probability. For the second question, you wish to find $P(\text{green} \mid\text{orange})$. Here Bayes' formula (and your answer from the first question) will be helpful. For the last question, you need to ...


1

Split up like $52=4+4+44$, i.e. $4$ kings $4$ aces and the other $44$ cards. In total $N:=\binom{52}2$ different pair-picks are possible and equiprobable. $\binom40\binom40\binom{44}2$ of them give $(K,A)=(0,0)$ $\binom40\binom41\binom{44}1$ of them give $(K,A)=(0,1)$ $\binom41\binom40\binom{44}1$ of them give $(K,A)=(1,0)$ $\binom40\binom42\binom{44}0$ ...


1

With the well deserved criticism of the problem in my Comments, let's see how one would solve it if we ignore some of the unnecessary clauses in the problem and we make it work by changing one of the inputs. Let's say only 60 customers buy cake at Cafo (to make the numbers work with the other conditions), and the problem doesn't say stupid things like "each ...


1

I think it must be proved that $\mu=\mathcal N(0,\sigma^2)$ but for convenience I will also preassume that $\sigma=1$ If $\phi$ denotes the characteristic function then:$$\phi(t)=\phi\left(\frac{t}{\sqrt2}\right)^2$$ Note that this can be repeated to arrive at $\phi(t)=\phi(\frac{t}2)^4$ and can be repeated again. Actually with this it can be shown that ...


-1

Hint: without relying on the CLT, you can use characteristic functions. Let $f$ be the characteristic function of the law $\mu$. You can easily show that $f$ satisfies this functional equation which can be solved to find that $f$ is the characteristic function of $\mathcal{N}(0,1)$.


0

You have that $X_k\sim\mathcal N(1.8\textrm{cm},0.7^2\textrm{cm}^2)$ independently for all $\{X_k\}_{k\in\Bbb N^+}$ being the widths of books to be placed on a shelf of length $4600$cm. Now, let $Y_N=\sum_{k=1}^N X_k$ , and so $Y_N \sim \mathcal N(1.8\textrm{cm}\,N, 0.7^2\textrm{cm}^2\,N)$ $Y_N$ is thus the combined width of $N$ books. You want to find ...


1

So, I will explain a solution that does not introduce a binomial random variable. Concretely, there is only one road that will lead him to the mall and three other roads that will not lead him to the mall, but back to the starting position. The problem is asking: what is the probability $A$ that in his first attempt, he chooses one of these three incorrect ...


1

$f(0)=e^{-\lambda}$, $f(1)=e^{-\lambda} \lambda$ , $f(2)=\frac{e^{-\lambda} \lambda^{2}}{2}$. Hence $2e^{-\lambda}+\frac{e^{-\lambda} \lambda^{2}}{2}=2e^{-\lambda} \lambda$, dividing by $e^{-\lambda}$, we get $\frac{\lambda^{2}}{2}-2\lambda+2=0$, hence $\lambda^{2}-4\lambda+4=0$, so $(\lambda-2)^{2}=0$ so $\lambda=2$


0

$$2f(0)+f(2)=2f(1)$$ $$e^{-\lambda}(2\lambda^0/0!+\lambda^2/2!)=e^{-\lambda}(2\lambda/1!)$$ $$2+\lambda^2/2=2\lambda$$ Solve the quadradic equation for $\lambda$ and you are done.


0

Presumably the traveler won’t choose the same road again on his second try, so this is sampling without replacement. For the second choice to be correct, the first has to be incorrect, so the probability is, assuming that the choices are uniformly distributed, $\frac34\cdot\frac13=\frac14$.


1

Call the one particular black ball we're concerned with $B$. Then the orders in question are: $$ \begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & ...


0

Maybe this helps: We know that: $0<x<1$, $0<y<1$, $Z=XY^2$ and $W=Y$ first for $Z$: $$0<y<1 \implies 0<y^2<1 \implies 0<xy^2<1 \implies 0<z<1$$ for $W$, just replace $y$ by $w$: $0<y<1 \implies 0<w<1$


1

let $\varphi(x) = \log(f(x))$ so $\varphi(x) +\varphi(y)=\varphi(p) +\varphi(q)=F (p, q)=F (x, y)$ The only way this invariance can happen under the transformations: $ x+y=p+q $ $\quad$ $x^2+y^2=p^2+q^2 $ $ \quad$ is if $\varphi(x) +\varphi(y)=F (p+q, p^2+q^2)=F (x+y, x^2+y^2)$ consider orthogonal vectors $x\cdot y=0$ and $q=0$ so that $p$ is completely ...


3

Hint: In other words, $Y = \lfloor X\rfloor$. Recall that $$\lfloor X \rfloor = k\iff k\leq X<k+1.$$ Hence $$\{Y = k\}\iff\{\lfloor X \rfloor = k\}\iff \{k\leq X<k+1\}.$$


0

We have been given: $$f_{X,Y}(x,y) = y^{-1}e^{-y} ~[0\leq x\leq y]$$ So then we know: $$\begin{align}f_Y(y) ~ = ~& \int_0^y y^{-1}e^{-y}~\operatorname d x~~[0\leq y]\\[1ex] =~& e^{-y}~[ 0\leq y]\end{align}$$ Then, by change of variables (chain rule, Jaccobian, etc.): $$\begin{align}f_{XY\mid Y}(z\mid y) ~=~ & \lvert \dfrac{\mathrm d ...


0

We say two numbers have the same order of magnitude of a number if the big one divided by the little one is less than 10. For example, 23 and 82 have the same order as their increment number can be found inside the numbers affiliated number by timing the 1-3 f magnitude, but 23 and 820 do not. -John C. Baez This is the definition of "Order of ...


0

A Markov chain is ergodic if it is not periodic. Specifically, a period $k$ of an irreducible chain is defined as $k=gcd\{n: Pr(X_n = i |X_0 =i ) >0 \}$. Since state 1 can move to it self, then the $gcd=1$, which means the chain is not periodic, and thus, ergodic. Moreover, if we begin with a stationary distribution upon the states, then this is an ...


0

We see that it is possible to get to state $1$ from states $2,3,$ and $4$ by using the path $$4 \rightarrow 3 \rightarrow 2 \rightarrow 1.$$ We can also transition from $1$ to $1$ With non-zero probability. there is non-zero probability of transitioning from $1$ to $2$ and we can follow the path $$4 \rightarrow 3 \rightarrow 2$$ to get from $4$ or $3$ to ...



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