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It seems your fundamental difficulty is with definitions of probability concepts, not so much with the process of integration. If $X \sim Unif(50, 150),$ then the density function $f_X(x)$ has three parts: (i) For $x < 50,\;f_X(x) = 0;\;$; (ii) for $50 \le x \le 150,\;f_X(x) = 1/100 = 0.01;$ and (iii) for $x > 150,\;f_X(x) = 0.$ If you want to ...

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On 1) Let $W:=X+Y$. Then: $$F_{W}\left(w\right)=P\left(X+Y\leq w\mid X=0\right)P\left(X=0\right)+P\left(X+Y\leq w\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}F_{Y}\left(w\right)+\frac{1}{2}F_{Y}\left(w-1\right)$$ Here $F_{Y}$ is well known to you and knowing CDF $F_{W}$ you can find PDF $f_{W}$. On 2) $X=0\Rightarrow XY=0$ so that $P\left\{ ... 0 Consider the solid obtained by deleting from the cylinder with base a unit circle and unit height, the inverted right cone with the base being the top of the cylinder. Note that the height of this solid over any point$x$in the unit circle base is exactly the distance between$x$and the center. So the required average is \begin{eqnarray}\frac{\text{Volume ... 1 If we put a uniform distribution over a unit circle, then the pdf of the distance$\rho$from the center is given by: $$f_\rho(x) = 2x\cdot\mathbb{1}_{[0,1]}(x) \tag{1}$$ hence the average distance is given by: $$\int_{0}^{1} 2x^2\,dx = \color{red}{\frac{2}{3}}.\tag{2}$$ 0 This looks the same question as What is the probability the max of one distribution is greater than the max of a different distribution? where my answer was: There are two cases: If$\max_i x_i \le 5$(which happens with probability$\left(\frac{5}{10}\right)^n$since each$x_i$must be less than$5$) then the conditional probability that ... 0 You need$X_i \ge -\theta$and$X_i \le \theta$for all$i$. This can be rewritten as$-X_i \le \theta$and$X_i \le \theta$or as$\max(-X_i,X_i) \le \theta$for all$i$. If$X_{(1)}$is the smallest of the$X_i$and$X_{(n)}$is the largest then this can be written as$\max(-X_{(1)},X_{(n)}) \le \theta$. So your likelihood function is in fact ... -1 As I understand all random variables are asymptotically independent only. I do not think sample to sample independent exist. Your issue is always true for sample size. 0$\theta$is the parameter to estimate, which corresponds to the upper bound of the$U(0,\theta)$. The observed samples are$x_1=2,\,x_2=4$, and$x_3=8$. The likelihood function to maximize is$\mathcal{L}(\theta|X) = \frac{1}{\theta^n}$with$X$corresponding to the observed values (a vector, really),$\theta$the upper margin of the interval for which this ... 0 What you are doing is wrong. You must find the likelihood function. What you found is$1/\theta^n$? so where is it defined? It is true that$X_n$is the maximum likelihood estimator because it maximizes the true likelihood function. How do you find it? Added: Your answer is actually on the right direction but as I mentioned it is missing a crucial point ... 0 There is no such thing as being "totally random". For example, consider two experiments. In the first, I draw a value uniformly from$[0,1]$. In the second, I draw it uniformly from$[-1,1]$. Which of these is "totally random"? Also, it is misleading to say that each value is equiprobable, since this is true in every continuous distribution – every ... 0 If$(X_n, Y_n)$are dependent for all finite$n$, but converges in distribution (or "weakly") to some random variable$(X,Y)$where$X$and$Y$is independent, then we would say that$(X_n, Y_n)$are asymptotically independent. 1 Does$f_{X, Y}(x, y) = 2x+2y-4xy$fit the bill? 2 A sketch of$y=(x-x^2)^2$will reveal that if$X\in (0;1)$then$Y\in (0;0.0625]$, and that the section$0\leq Y\leq y$will correspond to two intervals on the support of$X$.$\begin{align} \mathsf P(Y\leq y) & =\mathsf P(0\leq (X-X^2)^2\leq y) & :\text{for } y\in(0;1/16] \\ & = \mathsf P\left(\left\{0\leq X\leq \Box\right\}\cup \left\{\Box\leq ...

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As jameselmore points out in their comment, the Poisson distribution is more generally suitable for this kind of problem, where the arrival process is the juxtaposition of a large number of independent, individually rare occurrences (no bus arrives more than once). Additionally, one big appeal of the Poisson distribution is that it's "memoryless," meaning ...

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Isn't a good approach to calculate the average wait time if B arrives at 2pm (50% probability) and the average wait time if B doesn't arrive at 2pm (50% probability), then average these two average times out, in 50% - 50% proportion. If B arrives exactly at 2pm, then B's average wait will be 30 minutes. If B doesn't arrive exactly at 2pm, I believe the ...

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