Sunday, 11 September 2016

calculus - Expected value of a continuous random variable: interchanging the order of integration



I have come across a proof of the following in Ross's book on Probability -



For a non-negative continuous random variable Y with a probability density function $f_Y$
$$

\mathrm{E} [Y] = \int_0^\infty P[Y \geq y]dy
$$



The author proves it by using



$$
\int_0^\infty \int_y^\infty f_Y(x)dxdy = \int_0^\infty (\int_0^x dy) f_Y(x)dx
$$



He refers to it as "interchanging the order of integration".

I have studied a fair amount of Calculus from Apostol's books (Vol 1 & 2). But I still can't seem to provide a proof of this equation. How does one go about proving this last equation?


Answer



we have
\begin{align*}
\int_{[0,\infty)}\int_{[y,\infty)} f_Y(x)\; dx\, dy
&= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)f_Y(x)\;dx\,dy\\
&= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)f_Y(x)\;dy\,dx\\
&= \int_{[0,\infty)}\int_{[0,\infty)}\chi_{[y,\infty)}(x)\;dy\cdot f_Y(x)\;dx\\
&= \int_{[0,\infty)} \int_{[0,\infty)} \chi_{[0,x]}(y)\; dy\cdot f_Y(x)\; dx\\
&= \int_{[0,\infty)} xf_Y(x)\; dx\\

&= E(Y)
\end{align*}
where $\chi_A$ denotes the indicator function of a set $A$.


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