Is ln(E(x^n)) equal to E(ln(x^n))

  • Thread starter xuej1112
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  • #1
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is ln(E(x^n)) equal to E(ln(x^n)), if x is a continuous variable and x>=0???

Help me! Thanks!!!!!!!
 

Answers and Replies

  • #2
Gib Z
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This would more appropriately have been placed in the Calculus and Beyond homework help section, even if it isn't homework it is of that nature and level. General Math is more suited for general discussion of mathematical ideas.

Anyway, what is E? Is it the exponential function? How are the exponential function and natural logarithm related.
 
  • #3
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Sorry. E is the expectation
 
  • #4
Gib Z
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Oh ok. In that case, no it's not true. In fact, in general E(g(x)) is not equal to g(E(x)) which is due to the definition of E(x) as an integral with x as the integrand. This makes it inherently difficult for these nice properties to occur, as integrals don't have "nice" properties past linearity. So while E(aX+bY) = aE(X) + bE(Y), E(X^n) is not anything nice in terms of E(x).
 
  • #5
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thanks. if g is a concave function, can i say g(E(x))>=E(g(x)) ? (jensen's inequality)
 
  • #6
Gib Z
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It's the other way around, g(E(X)) is LESS or equal to E(g(X)), if g is convex (not concave).
 

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