How can we show this must hold knowing that x is an exponential distribution with mean = 4? (Last time I forgot to include what x was!)
$$E\left( \Bigl(\frac{2}{3} \Bigr)^x \ \middle| \ x \ge 6 \right) \ge \Bigl(\frac{2}{3} \Bigr)^{10}$$
I have no idea how to start this question. Any hint what this question is about? any specific type of inequality? lower bound of a conditional distribution?
Thanks.
From memorylessness of the exponential distribution the left-hand side is equal to $E[(2/3)^{X+6}]$ so it remains to show $$E[(2/3)^X] \ge (2/3)^4.$$
Direct approach. With $\lambda=1/4$, $$E[(2/3)^X] = \int_0^\infty (2/3)^x \lambda e^{-\lambda x} \, dx = \lambda \int_0^\infty e^{-(\lambda + \log(3/2))x} \, dx = \frac{\lambda}{\lambda + \log (3/2)} = \frac{1}{1+4\log(3/2)}$$ and you can check numerically that this is larger than $(2/3)^4$.
I may be overlooking a simpler approach for proving the inequality though.