Saturday, 26 November 2016

probability theory - Let X be a positive random variable with distribution function F. Show that E(X)=inti0nfty(1F(x))dx

Let X be a positive random variable with distribution function F. Show that E(X)=0(1F(x))dx



Attempt




0(1F(x))dx=0(1F(x)).1dx=x(1F(x))|0+0(dF(x)).x (integration by parts)



=0+E(X) where boundary term at is zero since F(x)1 as x



Is my proof correct?

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