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0

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 ... 1 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 ... 0 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.$$... 3 Hint: We will consider the situation in terms of minutes. We notice that that they will meet only if |X-Y|\leq 5/60, where X,Y is the time of arrival as a fraction of the hour. We notice that X,Y\overset{iid}{\sim} \text{unif(0,1)}. Then they problem becomes$$P(|X-Y| \leq 5/60).$$It will help to draw a picture. No integration required. You can do ... 2 Hint: If f_{X,Y} is the multivariate pdf then you want to solve the following integral$$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. 0 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 ... 0 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 ... 1 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. 1 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}$$since ... 1 By definition, L is the max of U and 1-U. The max of two numbers is less than something if and only each of the numbers is less than that something. By this argument, we've shown:$$\{L<t\} = \{U<t, 1-U<t\}.$$1 U is the break point which lies uniformly distributed on (0;1). L is the length of the longer side of the break. This is somewhere on [\tfrac 1 2; 1) . When the break point U is less than 1/2 the length of the longer stick is 1-U, and other wise it is U. So, for any \tfrac 1 2\leq l\leq 1, then the longer length being less than ... 1 Given: f:[0,1] \to \mathbf{R} strictly increasing and U uniformly distributed over [0,1]. U uniformly distributed over [0,1] means: \forall_{x \in [0,1]}: P(U \le x) = x. We say that U's cumulative distribution function is x \mapsto x. Let's now look at the cumulative distribution function of f(U): \forall_{x \in \mathbf{R}}: P(f(U) \le ... 1 X_1 X_2=0 so E[X_1 X_2]=0 and thus, since E[X_1]=E[X_2]=0, the covariance and correlation must be zero. But, for example, X_1=1 \implies X_2=0 so they are not independent 2 The pdf of X=(X_1,\,X_2) is$$ 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, ... 0 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. 0 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}): ... 0 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, ... 1 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 henceX=2.9+0.2U,\qquad Y=2.9+0.2R,\qquad Z=2.9+0.2S,$$where U, R and S are independent and uniform on (0,1), (-1,1) and (0,2) respectively. Consequence: E(\max(X,Y,Z))=2.9+0.2\cdot E(\max(U,R,S)). ... 1 The PDF of Y = X_1 + X_2 is the convolution of the PDFs of X_1 and X_2$$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 $$P(Z-Y< t)= \int P(Z-Y<t \lvert Y=y) P(Y=y) dy = \int P(Z<y+t) P(Y=y) dy$$ 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 ...

2

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: $$P(\mbox{A} \ge ... 3 If you solve X(2-X)<0.5 for X you get X>1+\frac{1}{\sqrt{2}} and X<1-\frac{1}{\sqrt{2}} The cdf of X is F(x)=\begin{cases}0, \ \ x<0 \\ \frac{x}{2}, \ \ \ 0\leq x\leq 2 \\ 1, \ \ x>0\end{cases} It is F(X>x)=1-F(X<x) Therefore you have to calculate F(1-\frac{1}{\sqrt{2}})+(1-F(1+\frac{1}{\sqrt{2}})) Remark The ... 2 To elaborate on the discussion in the comments: Indicator variables can be very helpful for problems like these. Accordingly, let X_i be the indicator variable for the i^{th} value. Thus X_i=1 if your draw of x elements gets one of value i, and X_i=0 otherwise. It is easy to compute E[X_i]...if p_i denotes the probability that the ... 0 Suppose we have n items of m types where n=km. We draw p items and ask about the expected value of the number of distinct items that appear. First compute the total count of possible configurations. The species here is$$\mathfrak{S}_{=m}(\mathfrak{P}_{\le k}(\mathcal{Z})).$$This gives the EGF$$G_0(z) = \left(\sum_{q=0}^k ...

0

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 ...

0

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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