E(X)/E(Y) and E(X/Y)
In this part we aim to discover the relationship between \(E[\frac{X}{Y}]\) and \(\frac{EX}{EY}\) under the assumption that
\(X,Y\) are independent
Since \(X,Y\) are independent, then \[ \begin{gather*} P(X\leq x, \frac{1}{Y}\leq y)\\ =P(X\leq x, Y\geq \frac{1}{y})\\ =P(X\leq x)P(Y\geq \frac{1}{y})\\ =P(X\leq x)P(\frac{1}{Y}\leq y)\\ \end{gather*} \] We know that event \(A,B\) are independent if and only if \(P(A,B)=P(A)P(B)\);
Thus \(X,\frac{1}{Y}\) are independent. This implies that \[ E[\frac{X}{Y}]=E(X)E(\frac{1}{Y}) \] Assume \(E[X],E[\frac{1}{Y}]\) are finite.
The function \(\frac{1}{y}\) is strictly convex over the domain \(y>0\). So if \(Y>0\) with prob \(1\), then by Jensen's inequality we have: \[ E[\frac{1}{Y}]\geq\frac{1}{E[Y]} \] with equality if and only if \(Var(Y)=0\) which is when \(Y\) is a constant.
Thus if \(X,Y\) are indepenent and if \(Y>0\) with prob \(1\) then
- \(E[X]=0\Rightarrow E[\frac{X}{Y}]=0=\frac{EX}{EY}\)
- \(E[X]>0\Rightarrow E[\frac{X}{Y}]\geq \frac{EX}{EY}\) with equality if and only if \(Var(Y)=0\)
- \(E[X]<0\Rightarrow E[\frac{X}{Y}]\leq \frac{EX}{EY}\) with equality if and only if \(Var(Y)=0\)