# Tag Info

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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