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NOTE: as the OP points out in the comments, this solution incorrectly assumes that ALL balls in $C$ are subsequently tossed over towards $D$. That case may still be relevant so I will leave this up for now and will modify it if I can incorporate the filter from $C$. Let $X_i$ denote the indicator variable for the $i^{th}$ red ball. Thus $X_i=1$ if $r_i$ ...


2

Suppose $W = X$. Then $E[X\mid W] = X$ whereas $E[Y\mid W] = E[Y]$. There is too much freedom in the choice of $W$. You have to control for some property to arrive at a useful conclusion. Otherwise, any conclusion is possible as the example above shows.


2

$a$ is determined by the equation: $$\int_{-\infty}^\infty f(x) \ dx = 1$$So for this problem:$$a\int_{0}^\infty e^{-2x}-e^{-3x}\ dx = 1$$Now you can solve for $a$. For the second question, $$P(X\le 1)=\int_{0}^1 f(x) \ dx$$ Plug the values in and you can solve for the probability.


1

Hint: $a)$ to be a density function, it must be : $\displaystyle \int_{-\infty}^{\infty} f(x)dx = 1$. Split the integral. $b)$ $P(X\leq 1) = \displaystyle \int_{-\infty}^1 f(x)dx= \displaystyle \int_{0}^1 f(x)dx$


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Comment: It looks as if you may be using an inverse gamma prior distribution for Bayesian inference. See Wikipedia on 'inverse gamma distribution' for details. The inverse chi-squared distribution is a special case (just as 'regular' chi-squared is a special case of 'regular' gamma). Inverse gamma distributions are often used in Bayesian inference as ...


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I'd agree with Clarinetist. Consider a simpler example: Define $$ x + f(x) = c, $$ where $c$ is some constant. We cannot simplify even this one to some general form. You can see that if we let $f(x) = x^{2}$, we have $$ x + x^{2} = c $$ $$ \Rightarrow x = \frac{-1\pm \sqrt{1+4c}}{2} $$ which we cannot (without knowing $f$) define as some inverse function ...


0

The derivation of the Normal Distribution (Gaussian Distribution) should answer your question; it also explains why there is a pi and so forth: https://www.sonoma.edu/users/w/wilsonst/papers/Normal/default.html


1

The question, at least in the way you asked it, really doesn't make much sense, because that funny exponential formula is the definition of "normal". Your comments sound like you sort of don't get what I'm getting at here. An analogy: Suppose someone asked this: Q: How do you prove that a triangle has three sides? The answer would be this: A: Huh? Having ...


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We know for a continuous random variable, the PDF is the first derivative of the CDF wherever the CDF is differentiable. Here, as you see, the random variable is defined over the interval $ 1\le\ X \le\ 4 $ and the probability that it lies outside this interval is zero. Now, within this interval, the CDF is a piecewise function with only one point where ...


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Firstly, to have a continuous function $F$, you need to solve these equations which should be satisfied on the boundaries $F(1)=0.2(1)^2-0.4(1)+b=0$ $F(2)=0.4(2)+a=0.2(2)^2-0.4(2)+b$ Then, either you can find the corresponding PDF or use another formula to find the expected value, which works for positive random variables $\int\limits_0^{\infty} (1-F(t))...


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F(x)= 0 for t< 1 and $0.2t^2- 0.4t+ b$ for 1< t< 2. In order for the function to be continuous at t= 1, those must "match up": $F(1)= 0= 0.2(1^2)- 0.4(1)+ b= -0.2+ b$ so we must have $b= 0.2$. For 2< t< 4, F= 0.4t+ a. Again the two formulas must give the same value at t= 2. We must have $0=.2(2^2)- 0.4(2)+ 0.2= 0.4(2)+ a$. That reduces to ...


0

I have $c=\frac{2}{7}$ as well. The area is $0.5c+2c+c=1$ $3.5c=1 \Rightarrow c=\frac{2}{7}$ The pdf is $f(x)=\begin{cases} c\cdot (x-1), \ 1\leq x <2 \\ c , \quad \ 2\leq x <4 \\ c-1/2\cdot c\cdot (x-4), \ 2\leq x <4 \end{cases}$ The slope of the third interval is $\frac{f(x_1)-f(x_2)}{x_1-x_2}=\frac{c-0}{4-6}$ We can use the integral ...


0

Your density can be split in three parts $f_1$, $f_2$ and $f_3$: $f_1(x) \cdot x = -2/7 \cdot x + 2/7 \cdot x^2$ $f_2(x) \cdot x = 2/7 \cdot x$ $f_3(x) \cdot x = 6/7\cdot x -1/7 \cdot x^2$ Then then antiderivatives are: $I_1(x) = -2/14\cdot x^2 + 2/21 \cdot x^3$ $I_2(x) = 2/14 \cdot x^2$ $I_3(x) = 6/14 \cdot x^2 -1/21 \cdot x^3$ And finally the sum ...


1

Hint: $P(X=1|X\geq 1)=\frac{P(X=1 \cap X\geq 1)}{P(X\geq 1)}=\frac{P(X=1) }{P(X\geq 1)}$ and $P(X\geq 1)=1-P(X=0)$


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Just as another remark: a neat corollary which follows from this statement and the strong law of large numbers is that whenever $X_i$ is an iid sequence of r.v. such that the SLLN holds (e.g. $X_i \in L^1$ by Kolmogorov's SLLN), then $\mathbb{E}(X_1 \mid S_n) \to \mathbb{E}(X_1)$ $\mathbb{P}$-almost surely.


1

We need only show that, for any Borel set $A \in \mathbb{R}$, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} We denote by $F$ the common cumulative distribution function of $Z_1$ and $Z_2$. Then, from the independence assumption, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \...


1

Conditioned on the number of roots being $k$, the roots are uniformly distributed over all multi-sets of $\mathbb{F}_p$ of size $k$. Let $P(x)$ be a uniformly chosen polynomial chosen from $$ \mathbb{P} := \{f(x) = x^d + a_1x^{d-1}+\ldots + a_{p-1} x + a_p: a_i \in \mathbb{F}_p, f(x)\text{ has at least one root in }\mathbb{F}_p\}. $$ Each polynomial of $\...


1

Think it in the reverse way; Say your field contains $n$ elements. Choose uniformly any number of value from the field, say $p_1,p_2,\cdots , p_k|k\le n$ It is sure that $P(x)=\prod\limits_{i=1}^k(x-p_i)\in\Bbb{F}[x]$


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Let $(Z_1,Z_2)$ be uniformly distributed on $\{(-1,0),(0,1),(1,-1)\}$. Then $Z_1$ and $Z_2$ are identically distributed but dependent. The value of $Z_1+Z_2$ fixes the (different) values of $Z_1$ and $Z_2$, so $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$.


2

In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$. Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of $\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \...


1

The probability that $S_2$ is disjoint from $S_1$ is $\frac{\binom{p-d}{d}}{\binom{p}{d}} = \prod_{j=0}^{d-1} \frac{p-d-j}{p-j}$. This product is at least $\left( 1 - \frac{d}{p-d} \right)^d \ge 1 - \frac{d^2}{p-d}$. If $d = o( \sqrt{p})$, this is $1 - o(1)$, and hence the sets are almost surely disjoint. On the other hand, the product is at most $\left( ...


1

Let $m:=\min(n_1,n_2)$ and for $i\in\{1,\dots,m\}$ let $X_i$ take value $1$ if a red and a blue ball both with number $i$ will end up in box $C$. If this does not happen then let $X_i$ take value $0$. Then $$X:=X_1+\dots+X_m$$ equals the number of numbers doubly presented in box $C$. There are $2X$ balls that have a "numbermate" in the sense that there is ...


2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...


1

This is roughly a first order approximation for the distribution of velocities. Strictly speaking, the fraction of particles with velocities in the interval is $\int_{c_x}^{c_x+dc_x}f(c)dc$. But if $dc_x$ is small and $f$ is roughly continuous, then one can assume that $f$ is constant on the interval to obtain $$\int_{c_x}^{c_x+dc_x}f(c)dc\approx\int_{c_x}^{...


1

Integrating the joint density, we have \begin{align} \int\limits_{\mathbb R^2} f&=\int_0^\infty\int_{-y}^y C(y^2-x^2)e^{-y}\ \mathsf dx\ \mathsf dy\\ &= C\int_0^\infty \frac43 y^3 e^{-y}\ \mathsf dy\\ &= 8C, \end{align} so $C=\frac18$. To compute the marginal densities, we integrate the joint density. For $x>0$, $$ f_X(x) = \int_x^\infty \...


2

The change of passing no traffic lights at is $0.6$ The chance of passing exactly $1$ traffic light is: $0.4 \cdot 0.6$ - that because we have to pass the first light and stop at the second. following the same line, the chance of passing exactly $2$ traffic lights is: $0.4^2 \cdot 0.6$ The spacial case is passing exactly $5$ lights which is $0.4^5$ (we have ...


0

Since both the A and B measurements are of the same thing, why not just add the counts of A and B together to form a new distribution C? This assumes that both A and B's measurement methods are equally reliable, and that more measurements beget a more reliable estimation. If you don't have the counts for each value of X, but you have the number of ...


0

1-) The meaning of ∫P(Q)*dQ is the probability of finding the value of Q, according to the limits of the integration. 2-) The use of dR in the proof is fundamental to simplify the description of the shield of the sphere. Since we are talking about a sphere, nothing more appropriate than use spherical coordinates. In cartesian coordinates, it would be very ...


2

We get the same answer for any continuously distributed random variable. For let $M$ be a median (medians need not be unique). The probability that $X_i$ is $\gt M$ is $\frac{1}{2}$. The probability that the minimum is $\gt M$ is therefore $\left(\frac{1}{2}\right)^3$. Remark: Things can break down if the distribution is not continuous. For example, let us ...


1

You're right, this is no hard, although is quite tricky. The only thing you need is self-similarity. First note that $x^qa^{-q} = q\cdot\sup_{\lambda>0}(\lambda x - p^{-1}\lambda^p a^p)$. Then $$ \sup_{t\ge 0} \left(\frac{X_t}{1+t^{p/2}}\right)^q = q\cdot\sup_{t\ge 0}\sup_{\lambda>0}\left(\lambda X_t - p^{-1}\lambda^p (1+t^{p/2})^p\right)\\ = \sup_{\...


1

Okies, $U=Y-X \iff Y= X+U$ so we require the convolution: $$f_{Y-X}(u) ~=~ \int_\Bbb R f_{X,Y}(x,x+u)\operatorname d x$$ Now the $X,Y$ support is: $$(X,Y)\in\{(x,x+u):x>0, x+u>0, 2x+u<1\}$$ Rearranged this becomes: $$(X,Y)\in\Big\{(x,x+u): \big(-1{<}u{<}0, -u{<}x{<}\tfrac{(1-u)}2\big)\vee\big(0{\leqslant}u{<}1, 0{<}x{<}\tfrac{(...


1

Here's elegant solution to the problem, drawn from the idea that the median of three $\text{Uniform}(0,1)$ random variables follows a $\text{Beta}(2,2)$ distribution. We can use this without generating three uniform random variables directly. First off, a formula exists for the joint density of two order statistics within a random sample, as well as the ...


1

A more elegant computation can be performed to obtain the required inverse of the $\operatorname{Beta}(2,2)$ CDF: if $U \sim \operatorname{Uniform}(0,1)$, then $$X = \frac{1}{2} + \sin \left( \frac{1}{3} \tan^{-1} \frac{U-1/2}{\sqrt{U(1-U)}} \right) \sim \operatorname{Beta}(2,2).$$ Is this computationally efficient? No. Using Mathematica, the command ...


1

The CDF does have a manageable inverse. Consider: $$ x^2(3-2x)=u, \;x\in[0,1] $$ That is, you want to know where the CDF crosses the level $u$. We simply need a mapping that tells us what $x$ corresponds to a given $u\in[0,1]$, with the condition that the $x$ we found is in $[0,1]$. This is the same as finding the roots of $x^2(3-2x)-u$ for some fixed $u\...


0

Corrected answer for question (b): Following Mr. Graham Kemp's clarifications I retried the second part (b) and then realized when a child is getting chosen randomly, $P(T)$ is getting weighted by the number of children each family has. So if we follow the above solution method we'd have (in case (b)): Total number of children $= 1\cdot 30 + 2\cdot 50 +...


1

Could anyone please explain if the assumption that (a) and (b) are one and the same is correct and if not why not? It is not correct. Let us look at a simpler scenario. Take two families, one with three children and one with a single child.   Here there are four children: two with birth order 1, and one each with birth orders 2 and 3. Case 1: First ...


2

Define random variables $X_1, X_2,\ldots,X_n$ where $X_i$ equals $1$ if voter $i$ votes for Candidate A, and $-1$ otherwise; define $S_k=X_1 + \cdots + X_k$. Then $S_n>0$ means that Candidate A is the winner, and $S_{\alpha n}>0$ means Candidate A is leading after $\alpha n$ votes. Calculate for any $\alpha\in(0,1)$ that $S_{\alpha n}$ has mean $0$ and ...


1

You are close to a solution. Let $W=\max(X_1,X_2,X_3)$. You found that $$F_W(w)=\frac{w^3}{\theta^3}$$ (for $0\lt w\lt \theta$). I changed letters a little because $z$ might cause confusion with the random variable $Z$, which is $\frac{W}{\theta}$. Ultimately you want to find $E(Z)$, which is $\frac{1}{\theta}E(W)$. So let us find $E(W)$. You know the cdf ...


1

The natural object for the $M$-th power of the number of fixed points isn't a subset of $M$ fixed points put an $M$-tuple of fixed points. Classify the $M$-tuples according to the subsets of entries that are equal. The number of these types is the Bell number $B_M$. A straightforward calculation shows that each such type contributes $1$ to the expected ...


0

$X = 165+ 6 Z_1$ Where $Z_n$ are standard normal random variables. $Y = 175 + A Z_1 + B Z_2$ a component the depends on X and a component that is independent from X $A^2 + B^2 = 64$ the variance of Y $6A = cov(X,Y)\\ \rho = \frac {cov(X,Y)}{\sigma_x\sigma_y} = 0.5\\ \frac {6A}{48} = 0.5\\ A = 4\\ B = 4 \sqrt 3\\ Y = 175 + 4 Z_1 + 4 \sqrt 3 Z_2\\ \frac 43 ...


3

Well, there's not much leeway, so it's probably most efficient to count by hand: $8+8+4=20$, $8+7+5=20$, $8+6+6=20$, $7+7+6=20$ – that's it, $4$ ways.


0

It's a Poisson distribution with parameter $\lambda(1-p)$. No need to perform the summation; you're just reducing the rate by a factor of $1-p$.


1

You can prove that the answer is $1/2$ by induction. Denote with $E_n$ the event that after $n$ draws the sum is even. Base case: $n=1$. You draw $1$ number from $\{1,2\}$. Obviously, the probability that the sum is even, is equal to the probability that you draw $2$, is equal to $1/2$, i.e. $$P(E_1)=\frac12$$ Induction step. Assume this holds for $n$, i....


2

If a number can appear more than once, then this is easily seen to be $\frac12$. You can see this if you stop to catch your breath after you've drawn the first $n-1$ numbers. Then you realize that whether the total sum will become even or odd depends only on whether the last number (that you haven't drawn yet) is even or odd (the sum might end up having the ...


3

Consider $$\oint_C dz \frac{e^{i k z}}{(1+z^2)^{(1+\nu)/2}} $$ where $k \gt 0$ and $C$ is a semicircular contour in the upper half plane of radius $R$ with a detour up and down the imaginary axis from the circular arc with a small circular path around the branch point at $z=i$ of radius $\epsilon$. The contour integral is then equal to $$\int_{-R}^R dx \...


1

I checked the integration, and since BruceET checked the result numerically, chances are that it's correct.


1

The error function is defined by \begin{equation} \operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}\ dt\tag1 \end{equation} and \begin{equation} \Phi (z)=\frac{1}{\sqrt{2\pi}}\int_0^z e^{-x^2/2}\ dx\tag2 \end{equation} Now, setting $x=t\sqrt{2}$ to the equation $(2)$ then the domain $0<x<z$ changes to $0<t<\frac{z}{\sqrt{2}}$ \begin{...


1

You can alternatively use the Normal approximation to the Binomial Distribution. Where: $\mu=np=(20)(0.7)=14$, $\sigma= \sqrt{npq}\approx 2.05$ and use the following: $P(a \leq X \leq b) \approx \phi(\frac{b+ \frac{1}{2}-\mu}{\sigma})- \phi(\frac{a-\frac{1}{2}-mu}{\sigma})$ However since we are looking to find an exact result, i.e $P(X=14)$ we simply ...


0

Seems fine. Minor comment for the last line. $(-1)^2$ rather than $(-1^2)$.


2

Since you wrote you want to prove each option right or wrong and so far only heuristics have been offered, here are some ideas for proofs, roughly in order of increasing difficulty: For $3$) through $5$) use the binomial distribution. The probability of an even result is $\frac12$. Let $K$ be the number of even results. We have $P(K=k)=2^{-100}\binom{100}k$;...



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