Tuesday, 8 January 2019

probability - Expectation of non-negative random variable



Let X be a non-negative random variable. In a proof for E[X]=0P(X>t)dt from the answer of this question, we use Fubini for the middle quality. Why do we need X to be non-negative? We basically have a double integral over a function f(X,t) which is 1 if X>t and else 0. So this function is non-negative for any X not just for non-negative random variables X, thus we could use Fubini regardless. Where is the flaw in my reasoning?


Answer



We require X to be non-negative, because otherwise
X[0,X)1 dt,   so in general  E[X]E([0,X)1 dt).



For example, if for some ω0Ω we have that X(ω0)=5<0, then [0,X(ω0))= and [0,X(ω0))1 dt=0X(ω0), so we are in trouble if P(ω0)>0. Thus, X is required to be non-negative in order for the first equality to hold, not because of the Fubini theorem.



Not that for a non-positive X you get
\begin{equation}
E[X] = -E[-X] =- \int_0^{\infty} P (- X > t)\ dt= \int_{-\infty}^0 P (X \end{equation}


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