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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 $$\operatorname{E}[\max(X,Y)] = \sum_{x=0}^n \sum_{y=0}^n \max(x,y) \binom{n}{x} p^x (1-p)^{n-x} \binom{n}{y} p^y ... 4 To approach the problem via normal approximations, let's first think about normal distributions in general to understand the distribution of the maximum we assume that X_1,X_2 are independently distributed as normal variables with mean \mu and standard deviation \sigma. We see that$$P(\max (X_1,X_2)<\mu+t\sigma)=P(X_1<\mu+t\sigma)\times ...

4

The CDF of $Y= \sin(\pi X)$ is $$F_Y (y) = \mathbb{P} (Y \leq y) = \mathbb{P} (\sin(\pi X) \leq y) = 2 \, \mathbb{P} \left(0 \leq X \leq \frac{1}{\pi} \, \arcsin(y)\right) = \frac{2}{\pi} \, \arcsin(y)$$ Hence, the PDF is $$f_Y (y) = \frac{2}{\pi \sqrt{1-y^2}}$$

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 ...

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).

3

$$F_{Z_n}(t)=Pr(Z_n \leq t)=\begin{cases}0 & ,t\leq 0 \\ \frac{\lfloor tn \rfloor}{n} &, 0 \leq t \leq 1 \\1 &, t \geq 1\end{cases}$$ When $t \in (0,1)$, we need to count how many $i \in \left\{1, \ldots, n\right\}$ that satisfy $\frac{i}{n}\leq t$, that is $i \leq nt$ and $i$ is an integer. Clearly, the answer is $\lfloor tn \rfloor$. Hence the ...

3

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

3

An easy example meeting your requirements might be to define $L(n)=c$ constant with $0 \lt c \le 1$ and $$f_n=P(X=n)=c (n-1)^{-\alpha} - c n^{-\alpha} \text{ for } n \ge 2$$ $$f_1=P(X=1)=1-c.$$ A more well known example might be the zeta distribution with $$f_n=P(X=n)=\dfrac{n^{-\alpha-1}}{\zeta(\alpha+1)}$$ where $\zeta(x)$ is the Riemann zeta function - ...

2

Consider $([0,1],\mathscr F ($borel sets on [0,1]$), m($lebesgue measure$))$ Let $X_n(x)=n1_{[0,\frac{1}{n}]}(x)$. This R.V. converges to 0 a.s. (it only does not converge to 0 for x=0) but $E(X_n(x)) = 1$ for every $n$ so $E(X_n) \not \rightarrow E(X) =0$ Let $X_1=1_{[0,\frac{1}{2}]}$, $X_2=1_{[\frac{1}{2},1]}$, $X_3=1_{[0,\frac{1}{3}]}$, ...

2

We have been given that: $$~f_{X_1,X_2}(s,t)= \tfrac 12 \mathbf 1_{s\in [0;1],t\in [0;2]}$$   Where $~\mathbf 1_{E}=\begin{cases}1 & : E\\ 0 & :\textsf{otherwise}\end{cases}~~$ is an indicator function of event $E$. Are you aware that the density of the sum is the convolution:? \begin{align}f_{X_1+X_2}(z)~=~&\int_\Bbb R f_{X_1,X_2}(s, ... 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 ... 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 ... 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. 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 ... 2 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.$$We assume that \varepsilon<1/2. By assumption, we can find a such that for each n,$$\mathbb E\left[X_n^2\mathbf 1\left\{\left|X_n\right|\gt a\right\}\right]\lt \varepsilon\mathbb ...

2

Below is an outline. The grey areas are spoilers — reveal them by placing your mouse over them if you want details, but I'd suggest you try by yourself before. Compute the expression of the CDF $F^{(n)}_X$ of $X\stackrel{\rm def}{=} \max(X_1,\dots, X_n)$, using independence. Recall that for $t\in\mathbb{R}$, $$F^{(n)}_X(t) = \mathbb{P}\{ X \leq t \} = ... 1 I agree with both your calculations and get \frac{1892}{23205} for both of them. 1 I think the first is correct (law of total variance), for the second I suggest law of total probability P(Y) =E[P(Y\mid X)]:$$P(Y\leq 1) = E[P(Y \leq 1\mid X)] = \int_{X} P(Y \leq 1 \mid x)\cdot f(x) dx$$Y\mid X = U \sim N(x,1). \ , We know that U <c, U \sim N(\mu,\sigma^2) \Leftrightarrow Z<\frac{c-\mu}{\sigma}, where Z \sim N(0,1). So: ... 1 Scenario B is equivalent to a scenario where each die has success probability \frac23 on the first roll and \frac16 on all further rolls, and is rerolled as long as it succeeds. We can use this to express the distribution as a single convolution, instead of the (x+y)-fold convolution that you wrote. The probability for k of m=x+y dice to succeed ... 1 A sanity check: verify you get the right expectation. (This is not sufficient, but at least it's a necessary condition for correctness). Since Y\sim U(X,1), we have$$\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)=&~ ...

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 ...

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 ...

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 ...

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 ...

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

$f(0)=e^{-\lambda}$, $f(1)=e^{-\lambda} \lambda$ , $f(2)=\frac{e^{-\lambda} \lambda^{2}}{2}$. 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 $\lambda=2$

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