# Tag Info

-1

Are you sure about the $564x^6$ term? All polynomial coeffs are divisible by $k$ with the exception of that one.

1

Well, at least the denominator seems to be $(x^2 + 3n^2)^{n+1}$...

1

$n= 100$, $\mu = 8 \times 10^5$, $\sigma = 3 \times 10^5$. Since sample size is more than $30$ we can use CLT and use normal approximation. a) $P(\bar{X} > 825000)$ = $P((\bar{X} - \mu)/ \sigma > (825000-800000)/300000)$ = $P(Z>1/12)$. Where $Z$ follows standard normal distribution. b) $P((7900000-800000)/300000 < Z < ... 0$ G(x)=Pr(X\leq x|a<X\leq b) = Pr(X\leq x, a<X\leq b) /Pr(a<X\leq b) = Pr(a<X\leq x) / Pr(a<X\leq b)$Now numerator is simplify as$ F(x)-F(a) $. Denominator is$F(b)-F(a)$. That's the answer. 0 There is no known distribution function but you can create by your own. Order the biased coin$1,2 \ldots ,M$. Now in the binomial coefficient we have$\binom Mk \times p^k \times q^{M-k}$. Here for a fixed$k$it will be$ (p_{i_1}p_{i_2}p_{i_3} \ldots p_{i_k})(q_{i_1}q_{i_2}q_{i_3} \ldots q_{i_k})$product will run over all possible choosing of$k$in ... 1 Suppose$Y$is exponentially distributed with expected value$\sigma$so that$\Pr(Y>y)=e^{-y/\sigma}, and let \begin{align} Z_1 & = \sqrt{-2\log_e Y}\,\cos X, \\ Z_2 & =\sqrt{-2\log_e Y}\,\sin X. \end{align} Then each ofZ_1,Z_2$is normally distributed with expected value$0$and variance$\sigma^2$and$Z_1,Z_2$are indepedent. See this ... 2 You may want to look for "Poisson binomial distribution", e.g., http://en.wikipedia.org/wiki/Poisson_binomial_distribution 0 We have: $$\begin{eqnarray*}\mathbb{P}\left[Y=\frac{Z}{\cos(X)}\in[0,\eta]\right]&=&\frac{1}{\sqrt{2}\,\sigma\,\pi^{3/2}}\int_{0}^{\eta}\int_{-1}^{1}\frac{e^{-\frac{(ux)^2}{2\sigma^2}}}{\sqrt{1-x^2}}\,dx\,du\\&=&\frac{1}{\sigma\sqrt{2\pi}}\int_{0}^{\eta}e^{-\frac{u^2}{4\sigma^2}}\,I_0\left(\frac{u^2}{4\sigma^2}\right)\,du\end{eqnarray*}$$ ... 1 Gamma(a) is nothing but a chi-square distribution. And you know t-distribution is ratio of Standard normal and chi-square distribution upon number of degree of freedom. From this relation you will get answer. 0$\left\{ Z>0\right\} =\cup_{n=1}^{\infty}\left\{ Z>n^{-1}\right\} $and consequently$\sum_{n=1}^{\infty}P\left\{ Z>n^{-1}\right\} \geq P\left\{ Z>0\right\} >0$. Then$P\left\{ Z>n^{-1}\right\} >0$must be true for some$n$. Taking$b>0$small enough (e.g.$b=\frac{1}{2}P\left\{ Z>n^{-1}\right\} $) we also have$P\left\{ ...

1

Suppose otherwise, that $P(Z>a)=0$ for all $a>0$. If your cdf is continuous, then $$0=\lim_{a \to 0}P(Z>a)=P(Z>0) \ne 0,$$ a contradiction.

0

Since $np > 10$ we can apply Central Limit Theorem and use the formula $\mathbb{P}\!\left(p < \hat{p}\right)= \mathbb{P}\!\left(Z<\frac{p-\hat{p}}{s}\right)$ where $s = \sqrt{\frac{p(1-p)}{n}}$ and $Z \sim N(0,1)$. My answer came out to be .6052 (I didn't use a standard normal table).

0

Let $f$ be the p.d.f. of $X$ (exponential distribution) and let $g$ be the p.d.f. of $Y$ (beta distribution). The c.d.f. of $W=XY$ is then: $$\int_0^1 \int_0^{w/y} f(x)g(y) \, dx \, dy$$ The bounds of integration come from the support of $X$ and $Y$ (i.e. X has a exponential distribution so $x \ge 0$, and $Y$ has a beta distribution so $0 \le y \le 1$).

1

Hint : Is there another possibility to all eight cases will be detected ; only one case will be missed ; two or more cases will be missed ?

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First of all, make sure that the probabilities add up to $1$: $0.1+0.1+0+0.2+0.4+0.2=1$. Good, so the data given to you represents a "complete system". Now, each event is independent, i.e., for every event, if it occurs then all other events do not occur. So the probability that $X \leq Y$ is simply the sum of the probabilities of the events in which $X ... 0 If$X$is a non-negative random variable, then it is not possible that$X \lt 0$. If$X$is a non-negative random variable and$E[X]=0$, then it is not possible that$X \lt E[X]\Pr(X=E[X])=\Pr(X=0)=1\Pr(X\le E[X])=\Pr(E[X] \le X)=1\Pr(X \gt E[X])=\Pr(X \gt 0)=0$However it might be possible that$X \gt 0$, even if with zero probability. For ... 1 As answered by Karl $$\int t^{\kappa } \exp{\left(-\rho t^{\alpha\kappa + 1}\right)} \, dt=-\frac{t^{\kappa +1} \left(\rho t^{\alpha \kappa +1}\right)^{-\frac{\kappa +1}{\alpha \kappa +1}} \Gamma \left(\frac{\kappa +1}{\alpha \kappa +1},t^{\alpha \kappa +1} \rho \right)}{\alpha \kappa +1}$$ where appears the incomplete gamma function. This can ... 1 Did you try the substitution$u=t^{\alpha\kappa+1}$? As far as I see, you get then something like$k\cdot u^{\beta}\exp(-\rho u)$($k,\beta$constants) as integrand, the integral is then similar to the incomplete Gamma function. https://en.wikipedia.org/wiki/Incomplete_gamma_function Hope that it helps. 1 We have that$X_1$and$X_2$are independent$\Gamma(\alpha_i,1)$random variables. Note that$X_1$and$X_2$are non-negative with values in$[0,\infty).$Given the transformation, the joint density of$Y_1$and$Y_2$is: $$f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= ... 1 Here How exactly are the beta and gamma distributions related? you can find in one of the solutions posted. X1/X1+X2 has betta distribution 1 For i=0,1,2,3,\dots, you have found that$$\Pr(Y=i+1)=e^{-i}(1-e^{-1}).$$Let k=i+1. You have found that for k=1,2,3,\dots, we have$$\Pr(Y=k)=(1-e^{-1})(e^{-1})^{k-1}.\tag{1}$$That's because e^{-i}=(e^{-1})^i=(e^{-1})^{k-1}. Formula (1) is exactly the formula for the probability that Y=k, where Y has geometric distribution with parameter ... 0 Let X \sim Exponential(1) with pdf f_x(x) and Y \sim PowerFunction(a,k) with pdf f_y(y): Then the pdf of Z = X Y is: where: The TransformProduct function is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors. ExpIntegralE[n,z] denotes the exponential integral function E_n(z)=\int ... 1 Since X and Y are independent, the joint density of X and Y is the product of the individual densities, so it is$$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$on the entire x-y plane. We want to find$$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$where D is the unit disk. Switch to polar coordinates. We want to find ... 0 to have A,B such that: A-B \sim U, both A,B must be linearly trasformed samples of the same uniform distribution: mini-proof:$$ \mbox{let: }u \sim U[0,1]\\ \begin{cases} s_1\leftarrow u\cdot \lambda_1+\mu_1\\ s_2\leftarrow u\cdot \lambda_2+\mu_2 \end{cases} \implies \\ s_1-s_2=u\lambda_1+\mu_1-(u \lambda_2+\mu_2) = ... 1 I would say$\Phi(\frac{c-\mu}\sigma)=0.08\Rightarrow \frac{c-\mu}\sigma =\Phi^{-1}(0.08)=-1.406\Phi(u)=\,\,$distribution function of the standard normal distribution$N(0,1)$You'll find in the tables N(0,1), in Ecxel (function "=NORMSINV(0,08)) etc. 2 Use the fact that the rv$X$defined as the number of children cured is distributed$Bin(20,0.9)$1 For every nonnegative integer$n$, the density$f$defined on$x\gt0$by$f(x)=\frac1{n!}x^n\mathrm e^{-x}$is the PDF of the gamma distribution$(n+1,1)$. Your case is when$n=3$. The parameter$n$could be any real number$n\gt-1$provided$n!$is replaced by$\Gamma(n+1)$. 0 You multiply them because you are looking for the probability of at least one left-handed woman AND at least one left-handed man. If you let A be the event that the sample contains at least 1 left-handed woman and let B be the event that the sample contains at least 1 left-handed man, then$Pr(A \cap B) = Pr(A) \times Pr(B)$0 One possible solution without mixing is to use averaging of parameters. For instance, if you have several normal distributions with means$\mu_1,\mu_2,...$and standard deviations$\sigma_1,\sigma_2,...$, you can calculate the average out of them and use$\bar{\mu}$and$\bar{\sigma}$for the super-distribution. Another possibility is to calculate average ... 1 Forgetting some irrelevant factorial and exponential terms, you seem to be trying to factor the ratio $$R=\frac{\theta^{u-v}\lambda^v}{(\theta+\lambda)^u}.$$ You are not very specific about the complete expression you get for$R$, but indeed, $$R=\left(\frac{\lambda}{\theta}\right)^v\left(\frac{\theta}{\theta+\lambda}\right)^u.$$ And$R$is also, as Casella ... 5 It is quite likely that you were told that$X$and$Y$are independent random variables, but neglected to pass on this information to us. Assuming that$X$and$Y$are independent Poisson random variables,$U = X+Y$is also a Poisson random variable with parameter$\theta+\lambda$. Thus, the probability that$U = m$($m$is a nonnegative integer here) is ... 1 Provided$n$is size of$\mathcal{X}$and$m$is size of$\mathcal{Y}$, all possible joint probability distribution functions$\mathcal{P}_{X,Y}$are$n+m$dimensional vectors with non-negative components such that sum or first$n$components and the other$m$components have to be one. If$n=m=2$, then$\mathcal{P}_{X,Y}$can be drawn as a square ... 2$\{n:n>0,X_n=X_0\}$is the set of non-zero indices ($n$) of random values ($X_n$) which are equal to$X_0$. The minimum value is the least such index; the first in the list.$\begin{align} \mathsf E[N] & = \mathsf E[\min\{n:n>0, X_n=X_0\}] \\ & = \sum_{n=1}^\infty n \mathsf P(n=\min\{n:n>0, X_n=X_0\}) \\ & = \sum_{n=1}^\infty n \mathsf ...

3

What does $N=\min\{n>0:X_n=X_0\}$ even mean? The set $S=\{n\gt0:X_n=X_0\}\subseteq\mathbb N$ is random, its minimum $N$ is a random variable. For example, if $(X_0(\omega),X_1(\omega),X_2(\omega),X_3(\omega),X_4(\omega))=(7,2,42,7,13)$ then $N(\omega)=3$.

0

If, as the caps hint at, you mean that $F$ and $G$ are cumulative distribution functions, the answer is yes. We need to verify that the product has the required properties. So we want to show that $F(x)G(x)$ is continuous from the right, that $\lim_{x\to\infty}F(x)G(x)=1$, that $F(x)G(x)$ is non-decreasing, that $\lim_{x\to -\infty} F(x)G(x)=0$. The ...

0

Hint: total probability is $1$.

0

It appears, the code is printing the distinct elements of the given 3 input set. The number of times 1 print happens = $k$ (all 3 inputs are the same). The number of times 2 prints happens = $3k(k-1)$ (two of the inputs are the same). The number of times 3 prints happens = $k(k-1)(k-2)$ (all the three inputs are different). So, the expected number of ...

0

$$M_{x2}(t) = \frac{6e^t}{t^2} + \frac{6}{t^2} + \frac{12e^t}{t} - \frac{12e^t}{t^3} + \frac{12}{t^3}$$ $M'_{x2}(t) = \frac{6e^t t^2}{t^4} - \frac{24 e^t t}{t^4} - \frac{12t}{t^4} + \frac{36 e^t}{t^4} - \frac{36}{t^4}$. By applying L'Hopital's rule at $t=0$, we get $E(X_{2}) = \frac{1}{2}$. $M''_{x2}(t) = \frac{6e^t}{t^2} - \frac{36 e^t}{t^3} + ... 1 You calculated the derivative incorrectly:$\frac{d}{d\rho} \log(1-\rho)=-\frac{1}{1-\rho}$. 0 Under condition$X_{r}=j$the distribution of$\left(X_{1},\dots,X_{r-1}\right)$is again multinomial. This with parameters$n-j$,$q_{1},\dots,q_{r-1}$where$q_{i}=\frac{p_{i}}{1-p_{r}}$. This leads to $$P\left\{ X_{1}=x_{1},\dots,X_{r-1}=x_{r-1}\mid X_{r}=j\right\} =\frac{\left(n-j\right)!}{x_{1}!\cdots x_{r-1}!}q_{1}^{x_{1}}\cdots q_{r-1}^{x_{r-1}}$$ ... 0 Now, for the numerator, I use the multinomial distribution, which gives $$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!} p_i^{x_i} p_r^j$$ Actually, using the multinomial distribution, one gets $$P(X_i=x_i \cap X_r=j) = \frac{n!}{x_i! j!\color{red}{(n-x_i-j)!}} p_i^{x_i} p_r^j\color{red}{(1-p_i-p_r)^{n-x_i-j}}.$$ Then one should get that the conditional ... 3 It is not totally obvious to me what you are asking for, but I would have thought the argument is something like $$\Pr(R \le \mu + \sigma \Phi^{-1}(u)) = \Pr(\epsilon \le \Phi^{-1}(u)) = \Phi(\Phi^{-1}(u)) =u$$ 1 The question is a mess and the proposed function is not an MGF. Fortunately, from what is hidden somewhere in the comments, one can deduce that the question is as follows: Consider$(X_1,X_2)$with density$f(x_1,x_2)=12x_1x_2(1-x_2)$on$0\lt x_1,x_2\lt1$. Find the MGF of$X_1$and$X_2$and deduce their means and variances. Well, if I had to solve ... 0 There's definitely something wrong with your MGF: it has a singularity as$t \to 0$. In fact $$M(t) = -12 t^{-2} + 7 + \dfrac{9}{2} t + \ldots$$ as can be shown most easily using Taylor series. So it is not a moment generating function. 5 Let$F(t)=e^{t/C}\int_t^\infty f(x)\,dx$. The stated assumption is equivalent to$F'(t)\le 0$for large$t$. Therefore,$\int_t^\infty f(x)\,dx=O(e^{-t/C})$. 0 Define the best case senario. I suspect the best case is just the one print call and just two compares. That the least number of function calls. If that is so then you want to find: the probability of choosing the same value three times from a set of$k$. For every value of Y there is$1/k$chance that X is the same value. Likewise for values of$Z$. ... 2 The probability you win a game is$\pi.$This is a random variable with a beta prior distribution with known parameters$(\alpha,\beta).$We observe the results of$n$games of which you win$w$and lose$\ell, n=w+\ell.$These games are assumed to be independent Bernoulli outcomes given the value of$\pi.$The posterior distribution of$\pi$is still beta ... 2 "Complete the square" to get the equivalent expression: $$f_{X,Y}(x,y)=\frac{1}{\pi \sqrt{2}}exp[-(x+y\sqrt{2}/2)^2]exp[-{y^2}/2]$$ Now for a fixed value of$y$the first exp term shows the conditional distribution of$X$is$N(-y\sqrt{2}/2,1/2).$If we integrate out$x$the first exp term disappears leaving only a constant and no$y$term. So the second ... 1 first find the cumulative density function for the random variable$X$: $$\mbox{pdf}(x)=4x^3 \implies \mbox{cdf}(x)=\int_{0}^{x}\mbox{pdf}(t)\mbox{dt}=x^4$$ now let's find the inverse of the cumulative density function: $$\mbox{cdf}(x)=x^4 \implies \mbox{icdf}(x)=\sqrt[4]{x}$$ let's sample a value in$\mbox{U}(0,1)$and use the inv. cum. density function ... 1 Carleman's condition holds for$X$hence the distribution of$X$is determined by its moments. Carleman's condition holds for$X^2$hence the distribution of$X^2$is determined by its moments. For every$n\geqslant1$, Krein's condition holds for$X^{2n+1}$hence the distribution of$X^{2n+1}$is not determined by its moments. The distribution of$X^{4}\$ ...

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