Wikipedia:Reference desk/Archives/Mathematics/2023 June 12

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June 12

Expected value as integral

For a non-professional, why exactly is the expected value defined as $\operatorname {E} [X]=\int _{-\infty }^{\infty }x\cdot f(x)dx$, as is stated here? Ideally, explain step by step please.--Hildeoc (talk) 22:09, 12 June 2023 (UTC)

If the probability density function of a random variable $X$ is given as function on the reals, we can approximate its expected value by partitioning the real number line $(-\infty ,+\infty )$ into an infinite collection of small intervals $[x_{i},x_{i{+}1}),i=-\infty ...\infty .$ The probability that an outcome of $X$ falls in the interval $[x_{i},x_{i{+}1})$ is $P\left(X\in [x_{i},x_{i{+}1})\right).$ Abbreviating this as $p_{i},$ the approximation is then given by the sum $\sum _{i=-\infty }^{\infty }x_{i}p_{i}.$

Now recall that $P\left(X\in [x_{i},x_{i{+}1})\right)=\int _{x_{i}}^{x_{i{+}1}}f(x)\,dx.$

You may see where this is going. The value of $\sum _{i}x_{i}p_{i}$ lies between the lower and upper Darboux sums for $\int _{-\infty }^{\infty }xf(x)\,dx$ for that same partitioning, so as the partitioning gets arbitrarily fine, the approximation tends to the value of that integral.  --Lambiam 05:14, 13 June 2023 (UTC)