# Bound on expectation of exponentially distributed random variables (RVs)

Let $X_A$ and $X_B$ be independent exponentially distributed RVs with means $\mu_A$ and $\mu_B$, respectively.

I have been running simulations that suggest $$\mathbb E \left[ \frac{X_B}{A X_A+BX_B+C} \right] \leq \mathbb E \left[ \frac{X_B}{A \mu_A+B\mu_B+C} \right],$$ where $A,B,C>0$ are constants.

My question is, does anyone no how I might go about proving this?

If this is true for every positive $(A,B,C)$, then, by continuity when $B\to0$, $$\mathbb E \left[ \frac{X_B}{A X_A+C} \right] \leqslant \mathbb E \left[ \frac{X_B}{A \mu_A+C} \right]=\frac{\mu_B}{A \mu_A+C}.$$ By independence and strict convexity of the function $x\mapsto1/(Ax+C)$ on $x\gt0$, the LHS is $$\mu_B\mathbb E \left[ \frac{1}{A X_A+C} \right]\gt\mu_B\frac{1}{A \mathbb E(X_A)+C}=\frac{\mu_B}{A\mu_A+C}.$$ Thus, the inequality cannot hold in full generality.
On the other hand, if $A\mu_A=B\mu_B$, then $AX_A$ and $BX_B$ are i.i.d. hence, by symmetry, $$2B\,\mathbb E \left[ \frac{X_B}{A X_A+BX_B+C} \right]=\mathbb E \left[ \frac{AX_A+BX_B}{A X_A+BX_B+C} \right]=1-C\,\mathbb E \left[ \frac{1}{A X_A+BX_B+C} \right],$$ and, by convexity, $$\mathbb E \left[ \frac{1}{A X_A+BX_B+C} \right]\gt\frac{1}{\mathbb E(A X_A+BX_B)+C}=\frac{1}{A\mu_A+B\mu_B+C},$$ hence, indeed, $$\mathbb E \left[ \frac{X_B}{A X_A+BX_B+C} \right]\leqslant\frac{\mu_B}{A\mu_A+B\mu_B+C}.$$