3
$\begingroup$

I need to calculate the average of the following quantity:

\begin{equation} S_n=\prod_{i=1}^nS(X_i) \tag{1} \label{eq:1} \end{equation}

with $S(X_i):=o_{X_i}b_{X_i}$, where each $X_i\in \mathcal{X}=\{1,2,\dots,m\}$ is drawn i.i.d. according to a probability mass function $\mathbf{p}=\{p_1,p_2,\dots,p_m\}$. The $o_X\in\{o_1,o_2\dots,o_m\}$ are fixed real numbers with each $o_j \geq 0$ and the $b_X\in \mathbf{b}=\{b_1,b_2\dots,b_m\}$ are also fixed real numbers with $b_j\geq 0$ and $\sum_{j=1}^mb_j=1$. A special case could be $\mathbf{b}=\mathbf{p}$.

The average I need can be written as follows, using equation (\ref{eq:1}):

\begin{equation} \bar{S_n}=\frac{1}{n} \sum_{k=1}^n S_k=\frac{1}{n} \sum_{k=1}^n \prod_{i=1}^k S(X_i) \tag{2} \end{equation}

Which can be rewritten as:

\begin{equation} \bar{S_n}=\frac{1}{n} \{S(X_1)+S(X_1)S(X_2)+\cdots + S(X_1)\cdots S(X_n)\} \tag{3} \end{equation}

Question. Is there a way to use the weak law of large numbers in order to write this average in terms of the probability mass function $\mathbf{p}$?

$\endgroup$
  • $\begingroup$ If I understand well the question, you would like to know the limit in probability of the sequence $\left(\bar{S_n}\right)_{n\geqslant 1}$, right? $\endgroup$ – Davide Giraudo May 26 '16 at 16:42
  • $\begingroup$ @DavideGiraudo That's right. $\endgroup$ – Ana S. H. May 27 '16 at 21:32
  • $\begingroup$ @DavideGiraudo The asymptotic behavior for large $n$ would be useful too. $\endgroup$ – Ana S. H. May 27 '16 at 21:44
1
$\begingroup$

It will depend on the parameters, as one could expect. For example, if $p_i=b_i=1/m$ and $o_i=1/b_i^2$, then $S(X_i)=m$ and the sequence $\left(\bar{S_n}\right)_{n\geqslant 1}$ goes to infinity.

However, if $\mathbb E\left[S\left(X_1\right)^2\right]\lt 1$, then $\bar{S_n}\to 0$ in probability (actually even in $\mathbb L^2$). To see this, expand the square and use independence of the sequence $\left(S(X_i)\right)_{i\geqslant 1}$.

$\endgroup$

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.