A&W assignment 4

T4.1 Recursive dice

a)

$\mathrm{Pr}[X=x,Y=y]=\mathrm{Pr}[X=x]\cdot \mathrm{Pr}[Y=y|X=x]=\frac{1}{6}\cdot \frac{1}{x}$ (for $1\le y\le x$) so

$$p_{X,Y}(x,y)=\{\begin{array}{l}\frac{1}{6x},1\le y\le x\le 6\\ 0,\text{otherwise}\end{array}$$

b)

Since $x\ge y$, we must have

$$\mathrm{Pr}[Y=y]=\sum _{x=y}^6\mathrm{Pr}[X=x,Y=y]=\sum _{x=y}^6\frac{1}{6x}$$

Therefore,

$$p_Y(y)=\{\begin{array}{l}\frac{1}{6}{\sum }_{x=y}^6\frac{1}{x},y=1,\dots ,6\\ 0,\text{otherwise}\end{array}$$

c)

calculate the expected value: $\mathbb{E}[X]=\frac{1+2+3+4+5+6}{6}=\frac{7}{2}$ $\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}\left[\frac{X+1}{2}\right]=\frac{\mathbb{E}[X]+1}{2}=\frac{\frac{7}{2}+1}{2}=\frac{9}{4}$ (using linearity of the expected value)

calculate the variance:

Since $Y|X=x$ is uniform on $\{1,\dots ,x\}$ we have

$$\mathbb{E}[Y^2|X=x]=\frac{1^2+2^2+\cdots +x^2}{x}=\frac{x(x+1)(2x+1)}{6x}=\frac{(x+1)(2x+1)}{6}$$ $$\mathbb{E}[Y^2]=\mathbb{E}[\mathbb{E}[Y^2|X]]=\frac{1}{6}\sum _{x=1}^6\frac{(x+1)(2x+1)}{6}=\frac{1}{36}\cdot \sum _{x=1}^{6}2x^2+3x+1$$ $$=\frac{1}{36}\left(2\sum _{x=1}^6{x}^2+3\sum _{x=1}^{6}x+6\right)=\frac{1}{36}(2\cdot 91+3\cdot 21+6)=\frac{251}{36}$$ $$\text{Var}[Y]=\mathbb{E}[Y^2]-\mathbb{E}[Y{]}^2=\frac{251}{36}-{\left(\frac{9}{4}\right)}^2=\frac{275}{144}$$