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0

You say those papers "appear" to use this result? The result is false - possibly they're not actually using it. Consider a function $f$ so that $f(x)=1-10\,|x|$ for $|x|\le1/10$, $\,f(x)=1/2$ for $2\le x\le 2+9/5$, and $f=0$ elsewhere. (Draw a picture!) You can easily verify that $f$ is unimodal but $d$ is not, if $k_1=0$ and $k_2=1$.


0

There's a systematic reason why the estimators don't work for zero variance. The mode is at $(\alpha-1)/(\alpha+\beta-2)$; you can use arbitrarily high values of $\alpha$ and $\beta$ to attain a given mode, and the higher you make them, the higher the probability density at the mode. So there's no pair of parameters with maximum likelihood in this case. ...


0

Because they model two different things. A Binomial Distribution, $\mathcal{Bin}(n, p)$, is that the count of successes in of $n$ Bernoulli trials with each trial having independent and identical probability of success $p$.   The count of such successes can range from $0$ to $n$. A Poisson Distribution, $\mathcal{Pois}(\lambda, \Delta t)$ is that of ...


2

$\textrm{Bin}(n,p)$ converges to $\mathcal{P}(\lambda)$ when $n\rightarrow +\infty$ and $np\rightarrow \lambda$. So these probabilities are only asymptotically equal.


0

I think the particular values of $m$ and $s$ you are using give rise to unexpected computational difficulties. Perhaps it is just too much to wish for two-place accuracy retrieving $s$ with a million simulated values. Also, taking the ordinary SD of lognormal data may not be the optimal way to estimate parameter $s$ (especially when it is small). I'm not ...


0

Your questions are quite simple to answer. In case of Chi-Square test, the p-value is computed as $1 - CDF_{\chi^{2}_{df}}(ts)$, where $CDF_{\chi^{2}_{df}}$ is a cumulative distributive function of Chi-Square distribution with $df$ degrees of freedom and $ts$ is the value of your test statistic. (Try to figure out why is p-value calculated this way - it is ...


1

$$D = b^2-4a = (4k)^2 - 4 \cdot 4 \cdot (k+2) = 16(k^2 - k - 2) = 16(k-2)(k+1)$$ We need $D \ge 0$ for real roots, which happens when $k \ge 2$ or $k \le -1$. Can you take it from here?


0

In the case of $n$ persons the graph of possible friendships is made of ${n \choose 2}$ pairs. For each edge $e =\{a,b\}$ of this graph we consider a random variable $X_e$ if $X_e = 1$ this means that they are friends. If $X_e=0$ they are not friends. ($a,b$ are two individuals of our population). We assume that $X_e$ are independent random variables and ...


0

$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$ $=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$ are independent Similarly, $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$ Now you can just do the calculation. I guess the final answer is 3/4


0

$\newcommand{\var}{\operatorname{var}}$The conditional distribution of $X_D-X_{\overline D}$ given the event $C=c$ is normal with expected value $\Big(\mu+g(c)\Big) - g(c) = \mu$ and variance $1+1=2$, since \begin{align} \var(X_D - X_{\overline D} \mid C=c) & = \var(X_D \mid C=c) + \var( -X_{\overline D} \mid C=c) \\[10pt] & = \var(X_D \mid C=c) + ...


1

Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ...


0

$\sin X$ and $\cos Y$ are identically distributed, with $\Pr(\sin X\le x)=\Pr(X\le\sin^{-1}(x))$ so the density of $\sin X$ is $\frac{d}{dx}\sin^{-1}(x)=\frac1{\sqrt{1-x^2}}$. Thus the product distribution is, for $-1\le z\le 1$, $$ f(z)=\int_{-1}^1\frac1{\sqrt{1-x^2}}\frac{1}{\sqrt{1-(z/x)^2}}\frac1{|x|}\,1_{[-x,x]}(z)\,dx $$ ...


0

If anyone was curious, I was able to solve this using optimization by minimizing the objective function $$ (F(b) - F(a) - 0.95)^2 + (F(a) - (1 - F(b))^2 $$ where $F$ is the cdf. The first term requires that $P(a \leq X \leq b) = 0.95$ and the second requires that $P(X < a) = P(X > b)$, thereby "centering" this $95\%$ interval on the median. MATLAB's ...


0

It suffices to use the following property of integrals $$f \leq g \Rightarrow \int f(y)\, dy \leq \int g(y)\, dy$$ and note that $$1_{\{0 \leq y <t\}} y^2 Q(y)\leq t^2 1_{\{0 \leq y <t\}} Q(y)$$ Analogous observations will allow you to solve the other inequalities.


0

I think that they are almost ok but it depends on what do you exactly need. I see two potential problems among the kernel functions you mentioned (they arise because kernel functions are made for density estimation but you want something little bit different). 1) All of these functions are symetrical around $0$ but you have distances that are always ...


2

I'll take the statement "All players are equally skilled." to mean that every player has a $50\%$ chance to win against every other player. $7$ players have been eliminated by $p_5$; the probability that $p_1$ was not among them is $24/31$. If $p_1$ has not been eliminated, the identity of the winner of $p_5$'s branch of the tournament is irrelevant to her; ...


0

First: Yes, the square root of the determinant of the $i$-th covariance matrix. Second: No, the inverse of the $i$-th covariance matrix. Yes, raising $2\pi$ to the $D/2$-th power. If $D$ is odd, this is a half integer; you still need to use $D/2$, without the rounding suggested in your code. $x^{D/2}=\sqrt{x^D}$.


3

Let $Y = E(X \mid G)$. Note that $$\int X^+ - X^-\, dP = \int X\, dP = \int Y\, dP = \int_{Y>0} Y\, dP + \int_{Y<0} Y \, dP = \int_{Y>0} X\, dP + \int_{Y<0} X \, dP $$ More over since $X$ and $Y$ have the same distribution $E [|X|] = E[|Y|]$. This implies $$\int X^+ + X^-\, dP = \int |X|\, dP = \int |Y|\, dP = \int_{Y>0} Y\, dP - ...


4

The simplest way is probably using the CDF, which is known in this case. Let $X \sim Exp(1)$ and $Y = \sqrt{X}.$ The CDF of $X$ is $F_X(x) = 1 - e^{-x}$, for $x > 0.$ Then $$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = 1 - e^{-y^2},$$ for $y > 0.$ Then take the derivative of the CDF of $Y$ to get $F_Y^\prime(y) = f_Y(y) = 2ye^{-y^2},$ ...


3

The median for a random variable $X$ is $m$ such that $P(X \le m) \ge 1/2$ and $P(X \ge m) \ge 1/2$. In the first example the correct answer is $0$: $P(X \le 0) = P(X = 0) = 0.728303$ and $P(X \ge 0) = 1$. In the second example it is $2$: $P(X \le 2) = 0.10 + 0.20 + 0.30 = 0.6$, $P(X \ge 2) = 0.30 + 0.25 + 0.15 = 0.7$. Your method is completely wrong.


1

Here's a table of payout probabilities that has the property that the expected return on a dollar is 96 cents, and which scales like 1/payout. With probability 71% (or so) you win nothing. Not sure if this is close to what you intend, but perhaps it is a start? 1: 0.137142857 2: 0.068571429 5: 0.027428571 6: 0.022857143 9: 0.015238095 10: ...


0

I would attempt to do a) and b) at the same time. By definition we have that $$F(x)=\int\limits_{-\infty}^x f(s)\;\mathrm{d}s=\int_0^x\frac{c}{\sqrt{s}}\;\mathrm{d}s=2c\sqrt{x}$$ where here $0\leq x\leq 4$. As has been mentioned in the comments, we now solve for $c$ using the fact that $F(x)\rightarrow 1$ as $x\rightarrow\infty$ by substituting $x=4$ into ...


0

As it turned out, my previous logic was wrong. Here is how it should be done. Marginal probability of choosing nest $k$ is $$P_{nB_k} = P\left[\max_{j\in B_k} U_{njk} \geq \max_{j\in B_s} U_{njs}, \forall s\neq k \right]\\ = P\left[W_{nk}+\epsilon_{nk}+\max_{j\in B_k}(Y_{nj}+e_{nj}) \geq W_{ns}+\epsilon_{ns}+\max_{j\in B_s}(Y_{nj}+e_{nj}), \forall s\neq k ...


0

Here $f(x) = \frac{1}{2\sigma} e^{-\frac{|x|}{\sigma}};\quad -\infty < x < \infty,\quad \sigma > 0$. Now, $$ \begin{eqnarray} E(X) &=& \frac{1}{2\sigma} \int_{-\infty}^{\infty} x e^{-\frac{|x|}{\sigma}} dx\\ &=& \frac{1}{2\sigma} \int_{-\infty}^{0} x e^{-\frac{|x|}{\sigma}} dx + \frac{1}{2\sigma} \int_{0}^{\infty} x ...


1

The probability an individual prisoner escapes on the first day is $0.4$. The probability an individual prisoner does not escapes on the first day is $0.6$. The probability an individual prisoner does not escapes on the first day but escapes on the second day is $0.24$. The probability an individual prisoner does not escapes on the first two days is ...


1

It's the factor $0.6^2$ that takes into account the possibility of prisoners escaping before the third day. On each day, the probability of a prisoner escaping is $0.4$, so the probability of the prisoner staying is $0.6$, and this is squared because there are two days before the third day on which the prisoner might escape.


3

Want to find $$ P(X \leq x \mid U \leq \mathrm e^{-X}) = \frac{P(X \leq x \cap U \leq \mathrm e^{-X})}{P(U \leq \mathrm e^{-X})}. $$ Note the joint density factors by independence: $f_{X,U}(x,u) = f_X(x)f_U(u) = \lambda \mathrm e^{-\lambda x} \cdot 1 = \mathrm e^{-x}$ with $\lambda = 1$. Numerator: $$ P(X \leq x \cap U \leq \mathrm e^{-X}) = \int_0^x ...


1

Comment, not answer: I did a simulation to try to break the problem into cases, and have no remarkable simplifications to offer beyond the revision of the suggestion by @AndreNicolas. In case it helps, with $T = X_1 + X_2 + X_3 + X_4,$ it seems $P(T \le 1) \approx 0.23.$ The same simulation shows $E(T) \approx 1.6$ and $P(T=0) \approx .0081,$ which are ...


2

Outline: The sum is $\le 1$ in several possible ways. (i) One of the $X_i$ is $1$ and the rest are $0$. The probability is $\binom{4}{1}(0.1)(0.3)^3$. (ii) There are four $0$'s. Easy. (iii) There is no $1$, and there are three $0$'s, (iii) There are no $1$'s, there are two $0$'s, and the sum of the remaining $2$ random variables in $\le 1$. Apart from ...


1

If $x$ is a real number then $e^x$ is a positive number less than $1$. The logarithm of a positive number less than $1$ is a negative number. $e^\text{something negative}$ is a positive number less than $1$. It has a logarithm. Its logarithm is negative.


1

The terminology may be a bit confusing, but when we say $Y$ is lognormally distributed, that means the logarithm of $Y$ is normally distributed. In other words if $X \sim \operatorname{Normal}(\mu,\sigma^2)$, then $\log Y = X$, and $Y = e^X \sim \operatorname{LogNormal}(\mu,\sigma^2)$. Then it becomes clear that if $X \in (-\infty, \infty)$, then $Y = e^X ...


0

When we write $e^X \sim logN$ we don't mean that we need to take the logarithm of the normal r.v. Rather, we mean that $X \sim \mathcal{N}(\mu, \sigma^2)$ and if we let $Y = e^X$, then $Y$ is log-normal. Note that since $X \in \mathbb{R}$, we must have $Y = e^X \in (0, +\infty)$. Now, note that since $Y$ is log-normal, we have $\ln Y \sim \mathcal{N}$, ...


2

I assume $X_1,X_2$ are independent. As Saty suggested, it's an easier transformation of variables if you set $$Y_1 = X_1+X_2\\ Y_2 = \dfrac{X_1}{X_1+X_2}.$$ Then $X_1 = Y_1Y_2$ and $X_2 = Y_1-Y_1Y_2$ and the Jacobian is $$ J = \begin{vmatrix} y_2 & 1-y_2 \\ y_1 & -y_1 \\ \end{vmatrix} = -y_1. $$ Then, ...


3

It sounds like you want to generate gamma-distributed random variables. If you want to write the code yourself, the way I know how is using the acceptance-rejection method by first generating an exponentially-distributed RV. An exponential random variable $G$ with pdf $\lambda \mathrm e^{-\lambda g}$ may be generated by first generating a uniform(0,1) ...


0

If $F$ is the pdf of the Gamma distribution with the correct parameters then for all real $x$, $$\Pr[\Delta X < x] = F(x).$$ This defines the distribution of $\Delta X$ completely.


0

In an answer to this question, I showed that if $X$ and $Y$ are independent zero-mean normal random variables with the same variance $\sigma^2$, then, with $\hat{X}$ and $\hat{Y}$ being random variables whose joint density is $4f_{X,Y}(x,y)\mathbf 1_{\{x>0, y> 0\}}$, $$F_{\hat{X}+\hat{Y}}(\alpha) = P\{\hat{X}+\hat{Y} \leq \alpha\} = ...


-1

Part b: \begin{align} \mathbb{P}(|X|=|Y|) &= \mathbb{P}(X=Y\ {\rm or}\ X=-Y) = \mathbb{P}(X=Y \cup X=-Y) = \mathbb{P}(X=XZ \cup X=-XZ) \\ & \hspace{5mm} = \mathbb{P}(X=XZ) + \mathbb{P}(X=-Y)-\mathbb{P}(X=XZ \cap X=-XZ) \\ & \hspace{10mm} = \mathbb{P}(Z=1)+\mathbb{P}(Z=-1)+0=\frac{1}{2}+\frac{1}{2} = 1 \end{align}.


2

The word "chance" is very loosely used in the question. Let's examine the case for 5 riders. If it is the expected value that is sought, E[x] = np, so for 5 riders, $5\cdot\frac15 = 1$ If it is the probability that is sought, P(none of the 5 have a puncture) = $(\frac{4}{5})^5 = \frac{1024}{3125}$ Thus P(at least one puncture occurs) = 1 - ...


1

This can be interpreted as a Bernoulli process. The probability of any single rider getting a puncture is (say) $1/5$. The probabilities of the different riders getting punctures are all independent, so to find the probability of certain people getting punctures, multiply $0.2$ for all the people who get punctures and $0.8$ for all the people who don't. To ...


1

Here is another piecewise linear solution: Write $$f(x,y)={1\over2}\bigl((1+g(x,y)\bigr)$$ whereby the function $g$ is defined by $$g(x,y):=\cases{x+y\quad&$(-x\leq y\leq 0)$ \cr x\quad&$(0\leq y\leq x)$ \cr}$$ in the sector $x\geq|y|$ and has the required symmetries. Here is a picture of the graph of $f$:


0

If there are $N$ riders, the probability of having no puncture is $$\Bbb{P}(\text{ no puncture }) = \bigg(\frac{4}{5}\bigg)^N $$ and that decreases with $N$. As the numbers of riders increase, the probability of occuring a puncture increases.


1

The simplest function I can come up with is the following: $$f(a,b)= \begin{cases} 0 & a + b \leq -1 \\ \frac{1+a+b}{2} & -1 < a + b < 1 \\ 1 & a + b \geq 1 \end{cases}$$ Here is a 3D graph of $z=f(x,y)$. Of course, it is not differentiable on the two lines where the function definition changes. Graphically, you can ...


0

\begin{align*} \mathbb P(X<Y)=&\,\underset{{\substack{x\in(15,45)\\y\in(0,60)\\x<y}}}\iint\left(\frac{1}{30}\times\frac{1}{60}\right)\mathrm d x\,\mathrm dy=\int_{x\in(15,45)}\left(\int_{y\in(x,60)}\frac{1}{1\mathrm,800}\,\mathrm dy\right)\mathrm dx\\ =&\,\int_{x\in(15,45)}(60-x)\times\frac{1}{1\mathrm,800}\,\mathrm d ...


0

I have been following this Question for about a day now, and am still not exactly sure what you are asking. Here are some comments based on sequences of one-minute time intervals, for which the number of $X$ of events in one minute is Poisson with mean 1/2. EXACTLY ONE event in such an interval: $P(X = 1) = 0.3032653.$ Geometric mean waiting time for the ...


0

My answer may be not rigorous but it's simple. I derive the joint density without cumulative function as follows: Let $\tau_{j}$ denote a permutation of $1,\dots,n$, and each permutation do not equals to each other. Thus, we have $$ \begin{align*} & f_{X_{(1)}, X_{(2)}, \dots, X_{(n)}}(y_1, \dots, y_n) \\ = & \Pr \{X_{\tau_1(1)} =y_1, \dots, ...


2

Let $C$ be the part of the plane where $x\lt y$. Then our probability is $$\iint_C f_{X,Y}(x,y)\,dx\,dy,$$ where $f_{X,Y}(x,y)$ is the joint density function of $X$ and $Y$. So the region of integration is the region above the line $y=x$. And yes, we assume that $X$ and $Y$ are independent. Express as an iterated integral. It is convenient to integrate ...


0

Just to get you on track: $P(X=1)=\frac23$ $P(X=2)=\frac13\frac24$ $P(X=3)=\frac13\frac24\frac25$ $P(X=4)=\frac13\frac24\frac35\frac26$ Do you see why? Do you recognize a pattern that enables you to find a closed form for $P(X=k)$? You are also asked to show that $\sum_{k=1}^{\infty}P(X=k)=1$ wich shows that you are dealing with a probability mass ...


0

\begin{align} \Pr(\text{W on 1st trial}) & = \frac 2 3 \\[10pt] \Pr(\text{W on 2nd trial}\mid \text{B on 1st}) & = \frac 2 4 \\[10pt] \Pr(\text{W on 3rd} \mid \text{B on 1st & 2nd}) & = \frac 2 5 \\[10pt] \Pr(\text{W on 4th} \mid \text{B on 1st 3}) & = \frac 2 6 \\[10pt] \Pr(\text{W on 5th} \mid \text{B on 1st 4}) & = \frac 2 7 \\ ...


2

The simple thing to do is to calculate small cases, detect a pattern, and generalize. For instance, the probability that you would only need to draw once is $$\Pr[X = 1] = \frac{2}{3},$$ because the initial state of the bucket is $2$ white and $1$ black. Thus the chance you get a white on the first draw is $2/3$, and the drawing stops upon the first draw of ...


1

Assuming $p(n)=0$ for $n<0$, $$p(n) = \frac{1}{n!}\left.\frac{d^n}{dx^n}F(x)\right|_{x=0}.$$ If your $F(x)=Ae^{Bx}x^C$, for $C$ a positive integer, then: $$F(x)=A\sum_{n\geq 0} \frac{B^nx^{n+C}}{n!},$$ giving $$p(n)=A\frac{B^{n-C}}{(n-C)!},$$ for $n\geq C$ and $p(n)=0$ otherwise.



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