Good morning,
Given an exponential distribution such that $X$ ~ $Exp(\lambda)$
with $Sn = \sum_{i=1}^n Xi$.
Given that $Cov(S_{31}, S_{57}) = 31 + 57$, find the lambda parameter of the exponential distribution.
I honestly do not know how to go about this. I know that the formula for Cov is : $Cov(X,Y) = E(XY) - E(X)E(Y)$ but this doesn't help me very much. I cannot determine $E(S_{31}S_{57})$ in my case.
Any help would be appreciated.
Suppose that $X_1,X_2,\ldots$ are exponentially distributed independent random variables with the parameter $\lambda$. If $n>m$, then $$ \operatorname{Cov}(S_m,S_n)=\operatorname{Cov}\biggl(\sum_{i=1}^mX_i,\sum_{i=1}^mX_i\biggr)+\operatorname{Cov}\biggl(\sum_{i=1}^mX_i,\sum_{i=m+1}^nX_i\biggr)=\operatorname{Var}S_m $$ and $$ \operatorname{Var}S_m=\sum_{i=1}^m\operatorname{Var}X_i=m\operatorname{Var}X_1 $$ using the independence and identical distributions. Since $\operatorname{Var}X_1=\lambda^{-2}$, $$ \operatorname{Cov}(S_m,S_n)=\frac m{\lambda^2}. $$ In this particular example $m=31$ and $\operatorname{Cov}(S_{31},S_{57})=31+57$. Hence, $$ \lambda=\sqrt{\frac{31}{88}}. $$