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alchimista

  • 2 years ago

Is there a random variable X such that E[X^2]=E[X]^2? if yes, is it self-independent ?

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  1. piscez.in
    • 2 years ago
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    this really seems like an interesting question, but maybe you can make it clearer, are you sure this is the right question?

  2. alchimista
    • 2 years ago
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    yes

  3. piscez.in
    • 2 years ago
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    is E[x} the exponential function

  4. alchimista
    • 2 years ago
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    i tought that Var(x)=E[X^2]-E[X]^2 , so for Var(X)=0 E[X^2]=E[X]^2

  5. alchimista
    • 2 years ago
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    no E[X] is the expectation of x

  6. piscez.in
    • 2 years ago
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    oh im sorry

  7. alchimista
    • 2 years ago
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    so based on my reasing then I have to find a random variable for which the Var(x)=0, i think

  8. alchimista
    • 2 years ago
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    ?

  9. paulstephen214
    • 2 years ago
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    if X is Cauchy Distributed then this will be true..., you can have the standardized Cauchy dist. 1/(pi*(1+(x^2)))

  10. Valpey
    • 2 years ago
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    Hard to call a variable "random" if it doesn't vary. It is also hard to show that this is true for the standardized Cauchy Distribution (basically a distribution with significant probability weight at (+/-)1/0).

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