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Mrrafil [7]
4 years ago
15

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]?

Mathematics
1 answer:
insens350 [35]4 years ago
5 0

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)}

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3 years ago
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