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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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If $R$ is a function of $wd$ then presumably $\mathbb{E}[aR]$ is not a constant, so its derivative of with respect to $d$ is not zero, with the consequence that your second and third lines are not helpful. To actually find optimal $d$, you will have to know what optimal means here, what the function $R$ is, and possibly some information about $a$. ...


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Let $\mathcal{U}_n=\{U_1, \ldots, U_n\}$ be a collection of iid random variables. Denote $U_{1:n} = \min \mathcal{U}_n$ and $U_{n:n} = \max \mathcal{U}_n$ be the smallest and the largest of the collection. Then $$\begin{eqnarray} F_{U_{1:n},U_{n:n}}\left(u_1, u_n\right) &=& \Pr\left(U_{1:n} \leqslant u_1, U_{n:n} \leqslant u_n\right) ...


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General approach for such a problem: There should be a formula in the relevant chapter that allows you to find the joint distribution of the max and the min. Transform to get the sum (twice the midrange) and difference (range). Integrate to find the density of the range. These formulas appear simple for uniform random variables, but be careful of the limits ...


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The average probability of a repair $R$ is 1/3 in one year. The probability of n repairs in the year is a Poisson distribution $$P(n) = \frac {R^n e^{-R}}{n!} $$ Assume each machine is independent, so the repair cost of two machines is twice the repair cost of one. The repair cost is uniformly distributed between 0 and 4000, so the expected average cost of ...


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The probability of $(0,0)$ occurring is $\frac23\cdot\frac23$; the probability of $(0,y)$ occurring, for $0<y\le4000$, is $\frac23\cdot\frac13\cdot\frac{dy}{4000}$ (and similarly for $(x,0)$); the probability of $(x,y)$ occurring, for $0<x,y\le 4000$, is $\frac13\cdot\frac13\cdot\frac{dx}{4000}\cdot\frac{dy}{4000}$. Therefore the expectation is $$ ...


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The expected waiting time $T_k$ to get down from $k$ to $0$ is $T_0 = 0$ for the base case, and otherwise it is $T_k = 1 + H_k$, where $H_k$ is the $k$th harmonic number $1 + 1/2 + 1/3 + \cdots + 1/k$. For large $k$ this is approximately $1 + \gamma + \ln k$, where $\gamma \doteq 0.57722$ is the Euler-Mascheroni constant. By inspection $T_0 = 0$. For $k ...


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I have no idea why this question is down-voted. It seems like a perfectly reasonable question to ask. The simplest thing I can think of is some kind of spinner with a pointer. Start the spinner and wait until it comes to a stop. The pointer then indicates the selected point on the spinner's rim. Not only is that point uniformly distributed on the ...


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Let $C$ and $D$ be the intersection points. Let $\theta$ be the angle $CAD$. The area of the diamond $ACBD$ is $\frac{1}{2}R^2\sin \theta + \frac{1}{2}R^2\sin \theta = R^2\sin \theta$. Twice the area of the sector $CAD$ of the circle is $R^2 \theta$. So the area shaded is $R^2(\theta - \sin \theta)$


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Probability of being within the shaded area = (shaded area)/(total area). Total area = $16R^2$. Shaded area - bit harder to work out! Copied from @user3491648's answer: Let $C$ and $D$ be the intersection points. Let $\theta$ be the angle $CAD$. The area of the diamond $ACBD$ is $\frac{1}{2}R^2\sin \theta + \frac{1}{2}R^2\sin \theta = R^2\sin \theta$. ...


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Bruce Trumbo's answer is good, but if you want a more basic approach then note that the probability of a single observation of being between $0.25$ and $0.75$ is $\frac12$, being low is $\frac14$, and being high is $\frac14$. The middle value will be in the central interval if any of the following patterns happen: all three observations central, with ...


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The limits of integration are changed because outside of those values the density is equal to zero. The random variable X is uniform on $(0, \sqrt{y})$ which means that outside of this range the density function is zero. That is the conditional density is, $ f(x \vert y) = \left\{ \begin{array}{lr} \frac{1}{\sqrt{y}} & : 0 < x < \sqrt{y} ...


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I don't know what you have covered so far, so I won't try to give you an analytic solution. You may get more help on that if you would edit your question to give some topics in the chapter preceding the problem, speculate on which ones might be related, and say what you think is keeping you from putting the pieces together. I will give you simulation ...


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The sum of discrete uniforms, taken to a quantity big enough is "normal" . Anyway as far as I know , even there's not theoretical distribution for what you're asking you alway can do this: If you have two variables X1, X2, then your space is all the possible values that X1+X2 can take, let's say that space goes from 1 to 26. Then for a value Z in that ...


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Suppose first to arrive has to wait for no longer than 5 minutes then the condition would of wait time would become 0 to 5 minute. Ram has to come from 12.15 to 12.45 Shyam has to come from 12.10 to 12.50 to maintain wait time no longer than 5 minutes As per the rule of uniform distribution P = 1/(b-a). f(X) dx So sample space would be time ...


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You already got a good answer to your specific question, but I would like to add a general remark. Yes, you can consider the joint distribution of a continuous r.v. $X$ and a discrete r.v. $Y$. One way to do it is to consider the joint CDF: $$ F_{XY}(x,y)=P(X\leq x,Y\leq y). $$ This is well defined for any two r.v. and you can compute marginal and ...


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If you expand the definition of expectation, you get $$ \mathbb{E} f(X,Y) = \int_{[0,1]} \sum_{y\in\{y_1,y_2\}} f(x,y)\mathbb{P}\{x\in dx\}\mathbb{P}\{Y=y\} = \int_{[0,1]} dx \left( f(x,y_1)\lambda + f(x,y_2)(1-\lambda) \right) $$ You can use a similar "return to the definition" to write the conditional expectations as well.


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Use change of variables: $$X = \sqrt{B^2-4C} $$ $$Y=C$$ $$B = \sqrt{X^2+4Y}$$ $$C=Y$$ Jacobian: $$ J = det\left( \begin{array}{ccc} \frac{\partial B}{\partial X} &\frac{\partial C}{\partial X} \\ \frac{\partial B}{\partial Y} & \frac{\partial C}{\partial Y} \\ \end{array} \right) = det \left( \begin{array}{ccc} \frac{X}{\sqrt{X^2+4Y}} & ...


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You want to find the mean of $\sqrt{B^2-4C}$. This cannot be done by substituting $E(B)$ and $E(C)$ for $B$ and $C$ in the expression $\sqrt{B^2-4C}$. Let $f_B(x)$ be the density function of $B$, and let $f_C(y)$ be the density function of $C$. So $f_B(x)=1$ on the interval $(2,3)$ and $0$ elsewhere, and $f_C(y)=1$ on the interval $(0,1)$ and $0$ ...


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Let's simplify to a single threshold $\theta \in [-a,a]$ and $f(x) = 0$ if $x<\theta$, $f(x)=1$ else. Suppose it is given that you sample $n$ times and you get $k_1$ samples to the left and $k_2$ samples to the right, where $k_1+k_2=n$ and $k_1\geq 1$, $k_2 \geq 1$. Let $Y_1$ be the max of those samples in the left region, and $Y_2$ is the min of those ...


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I will answer your question with a question: suppose you have a fair six-sided die and you roll it. You are interested in the probability of observing a value of at least $3$; e.g., if $X$ is a random variable that takes on values $X \in \{1, 2, 3, 4, 5, 6\}$, each with equal probability, then you are interested in the probability $\Pr[X \ge 3]$. Now, ...


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For the disc, $B$, $$f_{X|Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} = \dfrac{1/\pi}{2\sqrt{1-y^2}/\pi} = \dfrac{1}{2\sqrt{1-y^2}}\qquad\text{for $|x|\leq \sqrt{1-y^2},$ otherwise $0$.}$$ Symmetrically, $$f_{Y|X}(y|x) = \dfrac{1}{2\sqrt{1-x^2}}\qquad\text{for $|y|\leq \sqrt{1-x^2},$ otherwise $0$.}$$ So both these conditional distributions are uniform, which ...


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From your assumption it follows that $$\prod_{i=1}^n\bigl(p_i(1-p_i)\bigr)={1\over 4^n}\ .\tag{1}$$ If $p$ is a probability then $$p(1-p)={1\over4}-\left(p-{1\over2}\right)^2$$ implies that $p(1-p)\leq{1\over4}$ with equality only if $p={1\over2}$. Therefore $(1)$ can only hold if $$p_i={1\over2}\qquad(1\leq i\leq n)\ .$$


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I don't think so. $\begin{align} \rho & = \mathsf{Corr}(X,Y) \\[1ex] & = \dfrac{\mathsf E\Big(\big(X-\mathsf E(X)\big)\big(Y-\mathsf E(Y)\big)\Big)}{\mathsf E\Big(\big(X-\mathsf E(X)\big)^2\Big)^{1/2} \;\mathsf E\Big(\big(Y-\mathsf E(Y)\big)^2\Big)^{1/2}} \\[1ex] & =\frac{\mathsf ...


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Let $n=1$. The equation: $$\prod_{i=1}^1 p_i = \prod_{i=1}^1 (1-p_i)= \frac{1}{2}$$ does necessarily imply that $p_1=\frac{1}{2}$ is the unique solution. Let $p_i=\frac{1}{2}$ for all $i=1, \ldots, n-1$. Then by the same logic as the above: $$\prod_{i=1}^n p_i = \prod_{i=1}^n (1-p_i)= \frac{1}{2^n}$$ implies $p_n=\frac{1}{2}$ (divide off the product ...



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