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Note that $Var(X)=E(X^{2})-(E(X))^{2}$ thus since $$E(X)=\frac{1}{n+1}\sum_{i=0}^{n}i=\frac{1}{n+1}\frac{(n+1)n}{2}=\frac{n}{2}$$ and $$E(X^{2})=\frac{1}{n+1}\sum_{i=0}^{n}i^{2}=\frac{1}{n+1}\frac{n(n+1)(2n+1)}{6}=\frac{n(2n+1)}{6}$$ Thus we have $$Var(X)=\frac{n(2n+1)}{6}-\left(\frac{n}{2}\right)^{2}=\frac{n^{2}+2n}{12}=\frac{(n+1)^{2}-1}{12}$$ Thus the ...


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As in Jack's answer, there are $3$ cases, not $2$. $\frac{1}{2} \leq z \leq 1$: $$f_Z(z) = \int_{1/2}^{z} 2\cdot1 \,dy = \bigg[2y\bigg]_{1/2}^{z} = 2z-1 $$ $1 \leq z \leq \frac{3}{2}$: $$f_Z(z) = \int_{1/2}^{1} 2\cdot1 \,dy = \bigg[2y\bigg]_{1/2}^{1} = 2 - 1 =1 $$ $\frac{3}{2} \leq z \leq 2$: $$f_Z(z) = \int_{z-1}^{1} 2\cdot1 \,dy = ...


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The distribution for $0 \leq x \leq 2$ is $$P(|X| \leq x ) = \begin{cases} \int_{-x}^x\frac1{3}dt \,\, 0 \leq x \leq 1 \\ \int_{-1}^x\frac1{3}dt \,\, 1 < x \leq 2 \end{cases}= \begin{cases} \frac{2x}{3} \,\, 0 \leq x \leq 1 \\ \frac{x+1}{3}\,\, 1 < x \leq 2 \end{cases}.$$ Also $P(|X| \leq 0 ) = 0$ and $P(|X| \leq x ) = 1$ for $x > 2$. Taking the ...


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You have: $$f_Z(x)=(f_X*f_Y)(x)=\int_{-\infty}^{+\infty}2\cdot \mathbb{1}_{(1/2,1)}(y)\cdot\mathbb{1}_{(0,1)}(x-y)\,dy,$$ where $\mathbb{1}_J$ is the function that equals $1$ over $J$ and zero elsewhere. This can be written as: $$f_Z(x) = 2\int_{1/2}^{1}\mathbb{1}_{(0,1)}(x-y)dy $$ hence $f_Z$ is supported on ...


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$X \sim {\cal U}(0,1), f_{Y\mid X}(y \mid X=x) = \operatorname{\bf 1}_{(x, x+1)}(y) \operatorname{\bf 1}_{(0,1)}(x)$ a) What is the distribution of $Y$, given $X = x$? a) Is the uniform distribution just $(x,x+1)$? $\color{red}{\checkmark} \quad [Y\mid X=x] \sim {\cal U}(x,x+1), \forall x\in (0,1)$ b) What is $f(x,y)$? ...


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Answer is 1/6 ${(X+Y=2)}\cap{(X-Y=0)} = (X=1) \cap (Y=1)$ reason is : $(X-Y=0)\Longleftrightarrow (X=Y)$... so : $P((X+Y=2) \cap (X-Y=0))=P(X=1) \times P(Y=1)=(1/4)\times(1/4)=1/16$ (X and Y being independant) $P(X-Y=0)=P(X=Y)=P(((X=0) \cap (Y=0)) \cup ((X=1) \cap(Y=1)) \cup((X=2) \cap(Y=2)) = P((X=0) \cap (Y=0))+P((X=1) \cap(Y=1))+P((X=2) ...


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By definition $$P(X+Y=2 | X-Y = 0)= \frac{P(X+Y=2,X=Y)}{P(X=Y)} = \frac{P(X=Y=1)}{P(X=Y)} = \frac{1}{16 \cdot P(X=Y)}.$$ Now try to find $P(X=Y)$.



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