Die A has 4 red and 2 white faces whereas die B has 2 red and 4 white faces . A coin (fair) is tossed once . If it falls head , the game is carried on by throwing die A alone. If it falls tail die B is used . Let $X_n =1$ or $0$ according as the nth draw is white or red . and let $S_n = X_1 + X_2 +X _3 +\cdots+X_n$
now, i tried to find the expectation and variance of $S_n$ . and i think $S_n$ it follows Bin(n,1/2).
$P(X_n= 1)=P(X_n=1\mid \text{coin turned out to be head})\cdot0.5+P(X_n=1\mid \text{coin turned out to be tail})\cdot0.5= (4/6)\cdot(0.5) +(2/6)\cdot(0.5)=0.5$
so, $S_n $follows Bin (n, 0.5)]
and in that case , E($S_n$)=.5n and
Var($S_n$)=.25n----(1)
but if i solve it in this way :
Var($S_n$)=[Var($S_n$|coin turned out to be head)]*0.5+Var[($S_n$|coin turned out to be tail)]*0.5
=$[E(S_n|$coin turned out to be head$)-E(S_n)]^2$*0.5+$[E(S_n|$coin turned out to be tail$)-E(S_n)]^2$*0.5
implies Var($S_n$) =2n/9+$n^2$/36-----(2)
what is the reason of difference between the two results obtained in (1) and (2) ?
my book says $S_n$ doesn't follow binomial (n, 0.5) . is it the reason of this error ? but, why isn't it binomial ?
First off, note that a binomial distribution is for the sum of $n$ independent identically distributed Bernoulli variables. This is not the case here, since $\Pr(X_2=1|X_1=1)\ne\Pr(X_2=1|X_1=0)$, and hence we're not actually dealing with a binomial distribution. The conditional distributions of $S_n$, however, are binomially distributed, which we will use to our advantage. $S_n|Heads\sim Binom(n,\frac{2}{6})$ , and $ S_n|Tails\sim Binom(n,\frac{4}{6})$. Also, due to the symmetry of the problem, the expected value is still $E(S_n)=\frac{n}{2}$
Also, in your variance calculation, the law of total variance( https://en.wikipedia.org/wiki/Conditional_variance ) says that it is actually: $$Var(S_n)=Var(E(S_n|C))+E(Var(S_n|C)),$$ where $C$ represents whether the coin was heads($H$) or tails($T$). And so we need: $$E(Var(S_n|C)=Var(S_n|H)\cdot0.5+Var(S_n|T)\cdot0.5\\ =n(\frac{2}{6})(\frac{4}{6})\cdot0.5 + n(\frac{4}{6})(\frac{2}{6})\cdot0.5\\ =\frac{2n}{9}$$ and $$Var(E(S_n|C))=[E(S_n|H)−E(S_n)]^2\cdot0.5+[E(S_n|T)−E(S_n)]^2\cdot0.5\\ =(\frac{2n}{6}-\frac{n}{2})^2+(\frac{4n}{6}-\frac{n}{2})^2\\ =\frac{n^2}{18}$$ And so your variance should actually be $\frac{4n+n^2}{18}$