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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]?

User Kambala
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1 Answer

6 votes

Answer:


E(x^3) \geq 15625


E(√(x)) \leq 5


E(\log x) \leq 1.3979


E(e^(-x)) \geq e^(-(25))

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) = -√(x) is a convex function,


E(f(x))\geq f(E(x))\\\Rightarrow E(-√(x)) \geq -√((E(x)))\\\Rightarrow E(-√(x)) \geq -√((25))\\\Rightarrow E(-√(x)) \geq -5\\\Rightarrow E(√(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))

User Jeremy Rodi
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