Finding probability of three random variables and independence of two of them

Question

I am currently stuck on the below question(s). I've made some progress on part a, although I'm not sure if my reasoning is correct. It involved substituting the provided equation for C into the probability and then substituting M for m and moving stuff around to get 1/6. Again, I'm not sure if this is correct and any help is appreciated as I have no idea how to do the second part of the problem. Thank you

We are given two independent random variables, M and K, both taking values in {0, 1, 2, 3, 4, 5}. Further, assume that you know that the marginal distribution of K is uniform, i.e., $$P(K = i) = 1/6$$ for all i in {0,1,2...5} You do not know the marginal distribution of M. Finally, we define a new random variable C as $$C=M+K\: mod \:6$$

a) For every $$m,c \in {0,1,...5}$$ such that $$P(M=m) > 0$$ what is the probability $$P(C=c | M=m)$$ b) Show that M and C are (mutually) independent.

Answer 1

We use the notation $[K+M]_6$ to denote $K+M \text{ mod }6$ and compute for $c\in \{0,1,2,3,4,5\}$: \begin{align*} \mathbb{P}(C=c \: | M = m) &= \mathbb{P}([K+M]_6 = c \: | \: M=m) \\ &=\mathbb{P}([K+m]_6=c \: | \: M=m) \\ &=\mathbb{P}([K+m]_6 =c) \quad \quad \quad\quad\text{(by independence)} \\ &=\mathbb{P}(K = [c-m]_6) = \frac16 \end{align*} From here we can compute \begin{align*} \mathbb{P}(C=c) &= \sum_{m=0}^5 \mathbb{P}(C=c \: | \: M=m)\mathbb{P}(M=m) \\ &=\frac{1}{6}\sum_{m=0}^5 \mathbb{P}(M=m) \\ &= \frac{1}{6} \end{align*} And from the fact $\mathbb{P}(C=c \: | \: M=m) = \mathbb{P}(C=c)$ we can conclude independence of $C$ and $M$.

Answer 2

Since $K$ is uniformly distributed over $\{0,1,2,3,4,5\}$, so is $C$, independent of the value of $M$. Thus $P(C=c\mid M=m)=\frac16$. Independence then follows from $P(C=c\mid M=m)=P(C=c)$.

Answer 3

The event $(C=c)$ can be written as the disjoint union

$(C=c)=\Big[\bigcup_{m=0}^c(M=m,K=c-m)\Big] \bigcup \Big[\bigcup_{m=c+1}^5 (M=m,K=6+c-m)\Big]$

where we take empty unions to be $\emptyset$ (ie: when $c=5$). Now, taking probabilities, we have \begin{eqnarray*} P(C=c) &=& \sum_{m=0}^c P(M=m,K=c-m) + \sum_{m=c+1}^5 P(M=m,K=6+c-m)\\ &=& \sum_{m=0}^c P(M=m)P(K=c-m) + \sum_{m=c+1}^5 P(M=m)P(K=6+c-m)\\ &=& \sum_{m=0}^c P(M=m)\frac{1}{6} + \sum_{m=c+1}^5 P(M=m)\frac{1}{6}\\ &=& \frac{1}{6}\sum_{m=0}^5 P(M=m)\\ &=&\frac{1}{6} \end{eqnarray*} taking empty sums to be zero (ie: when $c=5$).

Now, partitioning the above on $(M=i)$, we can write the event $(C=c, M=i)$ as the disjoint union

$(C=c,M=i)=\Big[\bigcup_{m=0}^c(M=m,K=c-m,M=i)\Big] \bigcup \Big[\bigcup_{m=c+1}^5 (M=m,K=6+c-m,M=i)\Big]$

For $i\in \{0,\dots,c\}$ this reduces to $(C=c,M=i)=(M=i,K=c-i)$ and hence $P(C=c,M=i)=P(M=i,K=c-i)=P(M=i)P(K=c-i)=\frac{1}{6}P(M=i)$.

For $i\in \{c+1,\dots,5\}$ this reduces to $(C=c,M=i)=(M=i,K=6+c-i)$ and hence $P(C=c,M=i)=P(M=i,K=6+c-i)=P(M=i)P(K=6+c-i)=\frac{1}{6}P(M=i)$.

As a result, $$ P(C=c|M=i) = \frac{P(C=c,M=i)}{P(M=i)} = \frac{\frac{1}{6}P(M=i)}{P(M=i)}=\frac{1}{6} $$

Lastly, we note that $P(C=c)=P(C=c|M=i)=\frac{1}{6}$ showing that $C$ and $M$ are independent.