2
$\begingroup$

Let $X_n$ be a sequence of independent random variables with distribution: $$ P(X_i = -1)=q, \hspace{1cm} P(X_i = 0) = 1-2q = r, \hspace{1cm} P(X_i = 1) = q $$ for $i=1,..,n$ and $q \in (0,1/2)$.

I am investigating the distribution of $S_n = X_1 + X_2 + ... + X_n$.

What I have done for far:

let $p_n:= P(S_n = 0)$, then: \begin{equation} p_n = r^n + n(n-1)q^2 p_{n-2}. \end{equation}

Since all of the random variables $X_1, ..., X_n$ will be 0 or, the second possible option is that two of them will equal $1$ and $-1$ and the rest n-2 random variables will sum up to 0. Now we calculate $p_{n-2}$ the same way:

\begin{equation} p_{n-2} = r^{n-2} + (n-2)(n-3)q^2 p_{n-4}. \end{equation} (1) and (2) combined together give us: $$ p_n = r^n + n(n-1)q^2 p_{n-2} =r^n + n(n-1)q^2 [ r^{n-2} + (n-2)(n-3)q^2 p_{n-4}] = $$ $$ r^n + \frac{n!}{(n-2)!}q^2 r^{n-2} + \frac{n!}{(n-4)!}q^4 p_{n-4} = .... = r^n + \frac{n!}{(n-2)!}q^2 r^{n-2} + ... + \frac{n!}{(n-2k)!}q^{2k} r^{n-2k} + ... . $$

Analysing that we can deduce that:

$$p_n = P(S_n=0) = \left\{ \begin{array}{ll} \sum \limits_{k=0}^{n/2} \frac{n!}{(n-2k)!} q^{2k}r^{n-2k} & \textrm{when n is even},\\ \sum \limits_{k=0}^{\frac{n-1}{2}} \frac{n!}{(n-2k)!} q^{2k}r^{n-2k} & \textrm{when n is not even}.\\ \end{array} \right. $$

What seems to be accurate. I've checked for $n=1,2,3$ - so there is a chance that this calculation is correct.

Then I tried to calculate $P(S_n=i)$ as $$ P(S_n=i) = \binom{n}{i} q^i P(S_{n-i} = 0) $$ for $i>0$, as we need to have at least i ones (and we have to pick which $X_i$ are equal 1) and the rest (n-i) together must sum up to 0, and $$ P(S_n=i) = \binom{n}{-i} q^{-i} P(S_{n+i} = 0), $$ for $i<0$, but unfortunately this is incorrect.

I've checked in R that for $q=1/8$ and $n=3$: $$ \sum \limits_{i=-n}^{i=n} P(S_n = i) = 1.011719 \neq 1 $$

Please help me to find a mistake in my solution or help to calculate $P(S_n =i)$ in a different way.

$\endgroup$
  • $\begingroup$ Your recursion formula of $p_n$ is overcounting. Note e.g. that it tells you that $p_4=r^4+12q^2p_2=r^4+12q^2(r^2+2q^2)=r^4+12q^2r^2+24q^4$. The term $24q^4$ corresponds with possibilities without any zero's. But there are only $6$ of them (not $24$). $\endgroup$ – drhab Apr 8 '17 at 15:26
2
$\begingroup$

On base of non-negative integers $u,v,w$ with $u+v+w=n$ you could start with:

$$\Pr(U_n=u\wedge W_n=w)=\frac{n!}{u!v!w!}q^{u+w}r^v=\frac{n!}{u!(n-u-v)!w!}q^{u+w}r^{n-u-w}$$ where $U_n$ stands for the number of times that $-1$ and $W_n$ for the number of times that $1$ shows up.

Then: $$\Pr(S_n=i)=\sum_{w-u=i}\Pr(U_n=u\wedge W_n=w)$$


addendum:

If $B_n^{(i)}$ are iid for $i=1,2$ with Bernouilli($p$) distribution where $p$ is a root of the equality: $$p(1-p)=q$$ then $X_n$ and $B_n^{(1)}-B_n^{(2)}$ have the same distribution.

Defining $T_n^{(i)}=B_1^{(i)}+\cdots+B_n^{(i)}$ for $i=1,2$ we find that $S_n$ and $T_n^{(1)}-T_n^{(2)}$ have equal distribution, while $T_n^{(i)}$ are iid and have binomial distribution with parameters $n$ and $p$.

This observation can help by for instance finding things like the variance or characteristic function of $S_n$.

$\endgroup$
  • $\begingroup$ Thank you for your help! :) $\endgroup$ – Elizabeth_Banks Apr 8 '17 at 15:57
  • $\begingroup$ You are welcome. I added something. $\endgroup$ – drhab Apr 9 '17 at 13:33
  • $\begingroup$ Your help is much appreciated. $\endgroup$ – Elizabeth_Banks Apr 9 '17 at 14:24
2
$\begingroup$

I advise you to use generating functions. Let $Y_i=X_i+1$ (in order to work with positive valued random variables). In this way, because of independance, $S_n+n$ has generating function:

$$(q+rs+qs^2)^n=(s+q(1-s)^2)^n$$

It suffices to expand this expression in order to get the general term (don't forget to subtract $n$ at the end). In this way, one will find "trinomial coefficients" with some similarity with coefficients in the answer by @drhab.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.