Question

If Xi s a nonnegative random variable with mean 25, what can be said about:(a) E[X^3]?(b) E[√X]?(c) E[logX]?(d) E[e^−X]?

Answer 1

Answer:

$E(x^3) \geq 15625$

$E(\sqrt{x}) \leq 5$

$E(\log x) \leq 1.3979$

$E(e^{-x}) \geq e^{-(25)}$

Step-by-step explanation:

We are given the following information in the question:

$X_i$ s a nonnegative random variable with mean 25

$E(X) = 25$

We have to find the following:

a) Since $f(x) = x^3$ is a convex function,

$E(f(x))\geq f(E(x))\\\Rightarrow E(x^3) \geq (E(x))^3\\\Rightarrow E(x^3) \geq (25)^3\\\Rightarrow E(x^3) \geq 15625$

b) Since $f(x) = -\sqrt{x}$ is a convex function,

$E(f(x))\geq f(E(x))\\\Rightarrow E(-\sqrt{x}) \geq -\sqrt{(E(x))}\\\Rightarrow E(-\sqrt{x}) \geq -\sqrt{(25)}\\\Rightarrow E(-\sqrt{x}) \geq -5\\\Rightarrow E(\sqrt{x}) \leq 5$

c) Since $f(x) = -\log x$ is a convex function,

$E(f(x))\geq f(E(x))\\\Rightarrow E(-\log x) \geq -\log (E(x))\\\Rightarrow E(-\log x) \geq -\log (25)^3\\\Rightarrow E(\log x) \leq 1.3979$

d) Since $f(x) = e^{-x}$ is a convex function,

$E(f(x))\geq f(E(x))\\\Rightarrow E(e^{-x}) \geq e^{-(E(x))}\\\Rightarrow E(e^{-x}) \geq e^{-(25)}$