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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

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 - ...


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 ...


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) ...


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.


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 ...


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, ...


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 - ...


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.


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; ...


2

$P(F\mid E) = \dfrac{P(F\cap E)}{P(E)}=\dfrac{0.4}{0.8} = \dfrac{1}{2}$


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.


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 ...


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 $f$:


1

You can estimate the bounds by the $\min$ and the $\max$ of the sample. For the mode, several methods can be used. For example: the methods of moments (having previously estimated the two other parameters) estimate the density (with a kernel method) and then take the mode of the estimated density. chose a maximum likelihood estimator, as Chinny84 said.


1

An approach. For the triangle distribution you have three parameters $(a,b,c)$ where $$ P(X<a) = 0\\ P(X>b) = 0 $$ so you could use the $\min\lbrace X_i \rbrace$ and $\max\lbrace X_i \rbrace$ as pretty decent estimates. So the peak is the main thing to estimate, and that can be done with with the likeihood method. Do you know much about this ...


1

$\sin X$ 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 $-1\le z\le 1$, $$ f(z)=\int_{-1}^1\frac1{\sqrt{1-x^2}}\frac{1}{\sqrt{1-(z/x)^2}}\frac1{|x|}\,1_{[-x,x]}(z)\,dx $$ ...


1

Let $X$ be a standard $n$ dimensional Gaussian. Then $|X|$ is a random variable, and thus it has a distribution function. The author is saying let $F_1$ be that distribution function, i.e. $$ F_1(t) = P(|X| \leq t) $$ Similarly, the Euclidean ball of radius $r$ has finite volume and so we can consider the uniform measure $\mu$ on the ball (it will be a ...


1

We want the probability of not hitting to be $\lt 0.05$. The probability of not hitting in $x$ trials is $\left(\frac{1}{3}\right)^x$. We want this to be less than $0.05$. One does not really need theory to find the answer, just a bit of fooling around with the first few powers of $3$. But if we want to use logarithms, we have ...


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) = ...


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: ...


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?


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

There are $n-1$ edges available to person # $i$.   There is an independent probability $p$ of any of these being connected.   This description indicates it is a binomial distribution. $$\begin{align} \mathsf E(C_i) & = (n-1)\,p \\[2ex] \mathsf {Var}(C_i) & = (n-1)\,p\,(1-p) \end{align}$$ The interdependence of friend counts between ...


1

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 ...


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 ...



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