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I'll solve this in more generality, since question like this keep coming up and we need a standard answer to mark them as a duplicate of. So I'll calculate the probabilities for exactly $k$ bins to be empty and for at least $k$ bins to be empty when $m$ balls are distributed uniformly randomly over $n$ bins. Your case is $n=m=N$, with various $k$ given in ...


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It is because $0\leq y\leq x$ is the support if $Y$. When you factor in the support of $X$, the support of the joint distribution is , $0\leq y, \max(1,y)\leq x$ $$\begin{align}f_{Y}(y) =&~ \int_{y}^\infty f_{X,Y}(x,y)\operatorname d x~\big[0\leq y\big]\\[2ex] =&~ \int_{\max(1,y)}^\infty \tfrac 1{x^3}\operatorname d x~\big[0\leq ...


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In the general case, neither of these is right. You have two inequalities for $x$ in $f(x,y)$, namely $x\ge1$ and $x\ge y$, so the lower limit of the integral is $\max(1,y)$. You appear to be implying that the lower limit $y$ was specified in some book or lecture or the like. If so, I suspect that this was done because you need the value for $Y=\frac32$, and ...


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Since the distribution of $X$ is symmetric about $0$, we have $\mathbb E[X^n]=0$ for all odd $n$. For even $n$, we have $$\mathbb E[X^n] = \int_{-1}^1 \frac12 x^n\ \mathsf dx = \int_0^1 x^n\ \mathsf dx = \frac1{n+1}. $$ It follows that $$\mathbb E[XY]=\mathbb E[X^5]=0$$ and further $$\mathbb E[X]\mathbb E[Y] = \mathbb E[X]\mathbb E[X^4] = 0\cdot\frac15=0.$$ ...


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Hint: We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required. You can do ...


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Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$


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The change of variables $(uy,u)=(a,u)$ yields \begin{align} f_A(a) &= \int_{\mathbb R} f_U(u)f_Y\left(\frac au\right)\frac1{|u|}\ \mathsf du\\ &= \int_a^1 \frac1{u}\ \mathsf du\\ &= -\log a\mathsf \cdot1_{(0,1)}(a). \end{align} In general, if $X_1,\ldots,X_n$ are independent random variables and $\xi=\min_{1\leqslant i\leqslant n}X_i$, then a ...


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Assuming U,Y,Z are independent. For A: \begin{eqnarray} P(A>a) &=& P(UY>a) \end{eqnarray} The joint pdf of $U$ and $Y$ is $1$ withing the box $0\leq U \leq 1$ and $0\leq Y \leq 1.$ So compute the above probability by integrating 1 within the box $0\leq U \leq 1$ and $0\leq Y \leq 1,$ but above the curve $UY=a.$ You should get something like ...


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Hint: The density $f(a)=\frac12e^{-|a|}$ belongs to the Laplace distribution. The Laplace distribution arises when you subtract two iid exponential(1) variables. Second hint: If $U$ has uniform distribution on $[0,1]$ then $X:=-\ln (U)$ has exponential(1) distribution.


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A sanity check: verify you get the right expectation. (This is not sufficient, but at least it's a necessary condition for correctness). Since $Y\sim U(X,1)$, we have $$\mathbb{E}[Y\mid X] = \frac{1}{2}(1+X)$$ and therefore $$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\mid X]] = \frac{1}{2}(1+\mathbb{E}[X]) = \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$$ since ...


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By definition, $L$ is the max of $U$ and $1-U$. The max of two numbers is less than something if and only each of the numbers is less than that something. By this argument, we've shown: $$\{L<t\} = \{U<t, 1-U<t\}.$$


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$U$ is the break point which lies uniformly distributed on $(0;1)$. $L$ is the length of the longer side of the break.   This is somewhere on $[\tfrac 1 2; 1)$ . When the break point $U$ is less than $1/2$ the length of the longer stick is $1-U$, and other wise it is $U$. So, for any $\tfrac 1 2\leq l\leq 1$, then the longer length being less than ...


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Given: $f:[0,1] \to \mathbf{R}$ strictly increasing and $U$ uniformly distributed over $[0,1]$. $U$ uniformly distributed over $[0,1]$ means: $\forall_{x \in [0,1]}: P(U \le x) = x$. We say that U's cumulative distribution function is $x \mapsto x$. Let's now look at the cumulative distribution function of $f(U)$: $\forall_{x \in \mathbf{R}}: P(f(U) \le ...


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$X_1 X_2=0$ so $E[X_1 X_2]=0$ and thus, since $E[X_1]=E[X_2]=0$, the covariance and correlation must be zero. But, for example, $X_1=1 \implies X_2=0$ so they are not independent


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The pdf of $X=(X_1,\,X_2)$ is $$ f(x_1,x_2)=\begin{cases} \frac{1}{4} & \text{for } (x_1,x_2)\in Q=\{(-1,0), (1,0), (0,-1), (0,1)\}\\ 0 & \text{otherwise} \end{cases} $$ and the variable $X=(X_1,X_2)$ can be represented in tabular form $$ \begin{pmatrix} (X_1,X_2)\\ f(x_1,x_2) \end{pmatrix}= \begin{pmatrix} (-1,0) & (1,0) & (0,-1) & ...


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Hints: $|Y|=|XV|=|X||V|$ what can be said about $|V|$? $P(Y\in A)=P(Y\in A\mid V=-1)P(V=-1)+P(Y\in A\mid V=1)P(V=1)$ $P(Y\in A\mid V=v)=P(Xv\in A\mid V=v)=P(Xv\in A)$. The last equality because $X$ and $V$ are independent. $\mathbb EXV=\mathbb EX\mathbb EV$ again on base of independence. $\text{Var}(Y)=\mathbb EY^2-(\mathbb EY)^2$. Work that out. The PDF ...


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We have been given that: $$~f_{X_1,X_2}(s,t)= \tfrac 12 \mathbf 1_{s\in [0;1],t\in [0;2]}$$   Where $~\mathbf 1_{E}=\begin{cases}1 & : E\\ 0 & :\textsf{otherwise}\end{cases}~~$ is an indicator function of event $E$. Are you aware that the density of the sum is the convolution:? $$\begin{align}f_{X_1+X_2}(z)~=~&\int_\Bbb R f_{X_1,X_2}(s, ...


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An intuitive solution is to look at the unit square and consider the area of the upper triangle of $|X - Y| > 1/2$. Where $X$ is the arrival time of the one who has already arrived and $Y$ is the arrival time of the other person. The area and probability is $1/8$.


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Here’s another solution, basically the same as already given, but perhaps easier to follow, and at least more colorful. The ranges over which $\color{blue}X$, $\color{brown}Y$, and $\color{green}Z$ are uniformly distributed are shown above. Partition what can happen into the following three disjoint cases. Case 1 ($p=\frac{1}{2}$): ...


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Let $\ X,Y \tilde{} \mathcal{U} (\lbrack 0,1\rbrack) $ be uniformly distributed on $\lbrack0,1\rbrack$. 0 for 12.00, 1 for 13.00. Now we want to compute the probability that $X-Y \geq 0.5$. For this we will need to find the distribution function of the $Z=X-Y$. We can do this by convolution. For further details please check, ...


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First observation: The random variables $X$, $Y$ and $Z$ are independent and uniform on $[2.9,3.1]$, $[2.7,3.1]$ and $[2.9,3.3]$ respectively hence $$X=2.9+0.2U,\qquad Y=2.9+0.2R,\qquad Z=2.9+0.2S,$$ where $U$, $R$ and $S$ are independent and uniform on $(0,1)$, $(-1,1)$ and $(0,2)$ respectively. Consequence: $E(\max(X,Y,Z))=2.9+0.2\cdot E(\max(U,R,S))$. ...


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The PDF of $Y = X_1 + X_2$ is the convolution of the PDFs of $X_1$ and $X_2$ $$f_Y (y) = \begin{cases} \dfrac{y}{2} & \text{if } y \in [0,1]\\\\ \dfrac{1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3-y}{2} & \text{if } y \in [2,3]\\\\ 0 & \text{otherwise}\end{cases}$$ Integrating, we obtain the CDF $$F_Y (y) = \begin{cases} 0 & \text{if } ...


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This is more or less the definition of uniform distribution. Properties we certainly expect from a uniform (on $[0,1]$) random variable $X$ are that we want $\Bbb P([0,1])=1$ and $\Bbb P([a,b])=\Bbb P([a+c,b+c])$ whenever $0\le a\le b\le b+c\le 1$. Together with additivity, this already leads to $\Bbb P(X\in E)=\mu(E\cap[0,1])$.


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Hint: $f(\mathcal U) \le f(x)$ iff $\mathcal U \le x$ for $0 \le x \le 1$.


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have you tried this? any way I don't know if it works just hope it helps \begin{equation} P(Z-Y< t)= \int P(Z-Y<t \lvert Y=y) P(Y=y) dy = \int P(Z<y+t) P(Y=y) dy \end{equation} If you could find the distribution function of $Z-Y$ you would be able to determine it as a specific random variable. I mean you may want to work with distribution function ...


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Assume $X\sim U(0,2)$, that is the cut point is uniformly distributed between $0$ and $2$. Then if $X=x$ the area $A=a=x(2-x)$ and this is greater than a half between the roots of $x(2-x)=1/2$ which are $1\pm\sqrt{1/2}$. The pdf of $X$ is $f_X(x)=1/2$ for $x\in (0,2)$ and zero otherwise. So the probability that$A\ge 1/2$ is: $$ P(\mbox{A} \ge ...


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If you solve $X(2-X)<0.5$ for $X$ you get $X>1+\frac{1}{\sqrt{2}}$ and $X<1-\frac{1}{\sqrt{2}}$ The cdf of $X$ is $F(x)=\begin{cases}0, \ \ x<0 \\ \frac{x}{2}, \ \ \ 0\leq x\leq 2 \\ 1, \ \ x>0\end{cases}$ It is $F(X>x)=1-F(X<x)$ Therefore you have to calculate $F(1-\frac{1}{\sqrt{2}})+(1-F(1+\frac{1}{\sqrt{2}}))$ Remark The ...


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To elaborate on the discussion in the comments: Indicator variables can be very helpful for problems like these. Accordingly, let $X_i$ be the indicator variable for the $i^{th}$ value. Thus $X_i=1$ if your draw of $x$ elements gets one of value $i$, and $X_i=0$ otherwise. It is easy to compute $E[X_i]$...if $p_i$ denotes the probability that the ...


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Suppose we have $n$ items of $m$ types where $n=km.$ We draw $p$ items and ask about the expected value of the number of distinct items that appear. First compute the total count of possible configurations. The species here is $$\mathfrak{S}_{=m}(\mathfrak{P}_{\le k}(\mathcal{Z})).$$ This gives the EGF $$G_0(z) = \left(\sum_{q=0}^k ...


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If you are supposed to simulate $n = 50$ observations from a continuous uniform distribution on the interval $(2, 12),$ then make a histogram of the 50 observations and find their mean and variance, then you could do it it R statistical software as follows: x = runif(50, 2, 12) round(x, 2) # round to two places for a compact printout ## 8.22 11.47 ...


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I'll asume that you're drawing points uniformly from the rectangle $[a,b]\times[c,d]$ (which means that all the variables you mention are independent). We can't expect any kind of line-like structure to appear in the points you draw. $\alpha$ can be anything at all, and will be extremely sensitive to the actual points you have drawn. Since there is no reason ...


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It looks like you want to design a quantizer. A good textbook on this is "Vector Quantization and Signal Compression" by Gersho and Grey. One way to design such a mapping is to define a function (called a quantizer) $g(x)$ (which gives the approximated value) and then minimize some average of a loss function $L(x,g(x))$ (i.e. minimize $E[L(X,g(X))]$). ...



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