Say we have a sequence of random variables ( X n ) with mean μ n and variance σ n 2 , that converges in probability towards a random variable X. Can we reason that mean and variance sequences are bounded? Do we need some other assumptions, for example, E ( | X | ) &lt; ∞, or perhaps if we assume that each X n is normal? Thank you in advance.

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Say we have a sequence of random variables (      X    n  ) with mean       μ    n   and variance       σ    n    2  , that converges in probability towards a random variable   X. Can we reason that mean and variance sequences are bounded? Do we need some other assumptions, for example,   E  (      |    X      |    )  &amp;lt;  ∞, or perhaps if we assume that each       X    n   is normal? Thank you in advance.

Ryan Fitzgerald · Accepted Answer

Let   X be a non-negative r.v. with infinite expectation. Let       X    n    =  X      I          {      X      ≤      n      }       Then       X    n    →  X a.s., hence also in probability, but   E      X    n    →  ∞.If       X    n  &#039;s are normally distributed then   (      μ    n    ) and   (      σ    n    ) are convergent, hence bounded. [An easy way of proving this is to use characteristic functions:       E          i              μ        n            t            e          −              t                  2                            σ        n                  2                      →  E      e          i      t      X       and   E      e          i      t      X        ≠  0 if   t is positive and sufficiently close to 0 so we can take absolute values and then logarithms to finish].

dourtuntellorvl · Accepted Answer

thank you

Say we have a sequence of random variables ( X n </msub> ) with mean μ<!

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