Suppose $X \sim $ exp($\lambda$). Is it possible to find $P(X > 2E[X]/X > E[X])$ for any $\lambda$ in general? If possible, how should I proceed?
If not, how can I estimate the same probability value with a fixed $\lambda$?
I tried coding this problem and obtained different probability values for different $\lambda$ values. So, now I am interested in estimating the value mathematically using CDF or PDF.
If $$X \sim \operatorname{Exponential}(\lambda), \\ f_X(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then $$\operatorname{E}[X] = 1/\lambda$$ and $$\Pr\left[ X > \frac{2\operatorname{E}[X]}{X} > \operatorname{E}[X]\right] = \Pr\left[X > \frac{2}{\lambda X} > \frac{1}{\lambda} \right].$$ This compound inequality is equivalent to $$\frac{\sqrt{2}}{\sqrt{\lambda}} < X < 2,$$ hence $$\Pr[(2/\lambda)^{1/2} < X < 2] = \begin{cases} e^{-\sqrt{2\lambda}} - e^{-2\lambda}, & \lambda > 1/2, \\ 0, & 0 < \lambda \le 1/2. \end{cases}.$$