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June 12

Expected value as integral

For a non-professional, why exactly is the expected value defined as $\mathrm{E}[X]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\cdot f(x)dx$, 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 $(-\mathrm{\infty },+\mathrm{\infty })$ into an infinite collection of small intervals $[x_i,x_{i+1}),i=-\mathrm{\infty }...\mathrm{\infty }.$ The probability that an outcome of $X$ falls in the interval $[x_i,x_{i+1})$ is $P(X\in [x_i,x_{i+1})).$ Abbreviating this as $p_i,$ the approximation is then given by the sum $\sum _{i=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_i{p}_i.$

Now recall that $P(X\in [x_i,x_{i+1}))=\underset{x_i}{\overset{x_{i+1}}{\int }}f(x)dx.$

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 $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf(x)dx$ 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)