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The direction is not uniform Consider choosing points at random in a square in $2$ dimensional space. If they are uniformly distributed then the probability of choosing a given direction will be proportional to the number of points in that direction from the centre of the square. A line from the centre to a vertex will be $\sqrt{2}$ times longer than a line ...

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Yes, and in fact you don't have to assume that both densities are equal to $0$, just one of them. Let $[n]=\{1,\dots,n\}$. Since $$\frac{|(A\cup B)\cap[n]|}n=\frac{|A\cap[n]|}n+\frac{|B\cap[n]|}n-\frac{|(A\cap B)\cap[n]|}n$$ by the in-and-out principle, it follows that $\delta(A\cup B)$ exists whenever $\delta(A)$ and $\delta(B)$ and $\delta(A\cap B)$ all ...

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Yes. But note that it doesn't have to if you just assume that $\delta(A)$ and $\delta(B)$ exists. Indeed, let $A$ be the set of even integers and define $J: \mathbb{N} \rightarrow \mathbb{N}$ at an $n$ by letting $k$ be the unique integer such that $2^k\leq n<2^{k+1}$ and let $J(n)$ be $2n$ if $k$ is even and $2n-1$ if $k$ is odd. Putting $B:=J(\mathbb{N})... 3 Yes,$\delta(A \cup B)$exists and equals zero. Let$J_n = \{1,2,\dots,n\}$, and simply note that$|(A \cup B) \cap J_n| \le |A \cap J_n| + |B \cap J_n|$(it would be equality if$A,B$were disjoint). Hence $$\frac{|(A \cup B) \cap J_n|}{n} \le \frac{|A \cap J_n|}{n} + \frac{|B \cap J_n|}{n}.$$ As$n \to \infty$, both terms on the right side approach 0 by ... 0 The joint cdf of$Y$an$Z$is defined as $$F_{Y,Z}(y,z)=$$$$=P(V+W<y\cap V+X<z)=$$$$=\frac12\int_{-1}^{1}P(v+W<y\cap v+X<z\mid V=v)dv.$$ Since our random variables are independent, we may say that $$P(v+W<y\cap v+Y<z\mid V=v)=P(W<y-v)P(X<z-v).$$ The probabilities above can be calculated and then the integral with respect to$z$... 2 Note 1:$\dfrac{\mathrm d \sqrt y}{\mathrm d ~y~~}=\dfrac{1}{2\sqrt y}$Note 2:$x\mapsto x^2$folds$[-1;1]$onto$[0;1]$, so it effectively has two "inverse" functions. These are$y\mapsto +\surd y$and$y\mapsto -\surd y$, mapping$[0;1]$to$[0;1]$and$[-1;0]$respectively. Thusly: $$f_Y(y)=f_X({+}\surd y)\cdot\left\lvert\dfrac{\mathrm d ({+}\surd y)... 1 This is the probability (distribution function) what we have to evaluate:$$F_{X+Y}(u)=P(X+Y<u)=\iint_{A_u}\frac{1}{2} (x+y)e^{-(x+y)}dxdy$$where That is, we have$$\frac{1}{2}\int_0^u\left[\int_0^{-x+u} (x+y)e^{-(x+y)}dy\right]dx.$$Regarding the inner integral$$\int_0^{-x+u} (x+y)e^{-(x+y)}dy==u\sinh(u)+\sinh(u)-(u+1)\cosh(u)-x\sinh(x)-\... 1 I switch to lower case letters as arguments of the density functions, since upper case letters are usually reserved, in my experience, for the stochastic variables. As discussed in the comments if Y is a deterministic function of X and W than:$p(y|x,w)=\delta(y-f(x,w))$Therefore the integral becomes:$p(y|x)=\int dw \delta(y-f(x,w)) p(w|x)$Calling$...

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Let $f(x,y)$ be the mass distribution of the object (the density of the object at point $x,y$). Change the coordinates to polar coordinates: $$g(\rho,\theta) = \rho f(\rho \cos\theta, \rho \sin\theta).$$ Then the function $g(\rho,\theta)$ tells you the amount of mass of the object at distance $\rho$ and angle $\theta$. Finally, integrate over $\theta$ to ...

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It is $\int_x^{\infty} e^{-s}ds+\int_0^{x}\int_{x-s}^{\infty} \lambda e^{-\lambda x}dx e^{-s} ds$. [$s>x$ cannot be ignored in the computation].

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The integral in theory should be over $(-\infty,\infty)$ In practice this reduces to an integral over $[0,x]$ plus another over $(x, \infty)$ since $f_{E_1}(s)=0$ for $s \lt 0$ and $\Bbb P(E_0>x-s) = 1$ for $s \gt x$

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The domain of the joint density is the area above $|x|$ (red). So the marginal distribution is given by the following integral for any $x$: $$f_X(x)=\int_{|x|}^{\infty}\frac14 (y-x)e^{-y}\ d y=\frac14e^{-|x|}(|x|-x+1).$$

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If the $X_i$ are independent, then so are the $Y_i$, and $nZ$ is distributed as the sum of $n$ independent Bernouilli rvs, with expectations $p_i =P(a\le X_i\le b)=\Phi((b-\mu_i)/\sigma_i) -\Phi((a-\mu_i)/\sigma_i)$. The one number that tells the most is the sum $\lambda=\sum_{i=1}^n p_i$. In your applications, is it tiny, huge, or what? There is no ...

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You have to first find the cdf, differentiate it w.r.t $u$ to get the pdf. Let's start with the cdf computation. For $0\leq u\leq 2a$, \begin{align} F_U(u) &= \mathbb{P}(U\leq u)=\mathbb{P}(U^2\leq u^2)\\ &= \int_{ \left\{\substack{(x,y,z,t):-a\leq x,y,z,t \leq a\\ (x-y)^2+(z-t)^2\leq u^2}\right\}}f_X(x)f_Y(y)f_Z(z)f_T(t)dxdydzdt. \end{align} Here, ...

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No. If $$X,Y:(\Omega,\mathscr{A})\to (\mathbb{R},\mathscr{B}(\mathbb{R}))$$ are two random variables with $X,Y\in L^2(\mathbb P)$, then the covariance is $$\int_\Omega X(\omega)Y(\omega)\mathbb{P}(d\omega)-\mathbb{E}X\mathbb{E}Y=\int_\mathbb{R^2} xy ~\mathbb P_{(X,Y)}(d(x,y))-\mathbb{E}X\mathbb{E}Y$$ If $(X,Y)$ has a density $f_{(X,Y)}(x,y)$ with respect ...

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Try the change of variables $u=x+y, v = x$. Then $$\int^\infty_0\int^\infty_0\frac{f(x+y)}{x+y}\,dx\,dy=\int^\infty_0\int^u_0\frac{f(u)}{u} dv\,du=1$$

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