Wikipedia:Reference desk/Archives/Mathematics/2021 February 10

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February 10

Expected values

On the page of Expected value it is stated that if $X$ is a random variable defined on a probability space $(\Omega ,\Sigma ,\operatorname {P} )$, then the expected value of $X$, denoted by $\operatorname {E} [X]$, is defined as the Lebesgue integral $\operatorname {E} [X]=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega ).$. How does the definition for the case when $X$ is a random variable with a probability density function of $f(x)$, (in which case the expected value is defined as $\operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx$) follow from the general definition? Thanks - Abdul Muhsy (talk) 11:00, 10 February 2021 (UTC)

If a real-valued random variable has a probability density function $f$, it is the derivative of its cumulative distribution function $F$, so $\int _{S}y\,dF=\int _{S}yf(x)\,dx$ (see Probability density function § Absolutely continuous univariate distributions). The probability function $\operatorname {P}$ can then be equated with $F$ (see Random variable § Distribution functions). So $\int _{\mathbb {R} }x\,d\operatorname {P} =\int _{\mathbb {R} }x\,dF=\int _{\mathbb {R} }xf(x)\,dx$.  --Lambiam 15:20, 10 February 2021 (UTC)

Can you please explain the first equality in your last equation a bit more? Thanks Abdul Muhsy (talk) 18:03, 10 February 2021 (UTC)

If $\operatorname {P} =F$ (see the sentence immediately before the equation), their differentials are also the same.  --Lambiam 22:23, 10 February 2021 (UTC)

But I don't understand why $\operatorname {P} =F$ given that they have different domains (the sigma field and the real numbers respectively), and also why is $\operatorname {E} [X]=\int _{\Omega }X\,d\operatorname {P} =\int _{\mathbb {R} }x\,d\operatorname {P} .$! How does the integral on the probability space turn into an integral on the real line? Is it by first considering the pushforward measure, observing the expectations for X and the identity will be equal and then by Radon-Nikodym theorem applied for the pushforward measure and the Lebesgue measure? Which book will contain a detailed proof Abdul Muhsy (talk) 00:46, 11 February 2021 (UTC)

It is nothing deep. The event space ${\mathcal {F}}$ can be taken to be the Borel $\sigma$-algebra; then its distribution is a measure on ${\mathcal {F}}$ (see Borel set § Example). Given a probability distribution $F,$ the $F$-measure of an interval $I=[x_{0},x_{1}]$ is $F(I)=F(x_{1})-F(x_{0}),$ where the endpoints may be infinite. For an event ${\mathcal {I}}$ represented as a set of disjoint intervals, ${\textstyle \operatorname {P} ({\mathcal {I}})=\sum _{I\in {\mathcal {I}}}F(I).}$ We can also go the other way: $F(x)=\operatorname {P} (\{[-\infty ,x]\}),$ which shows that as far as probability distributions on $\mathbb {R}$ are concerned, there is a one-to-one correspondence between distributions and probability functions. The lower-case variable $x$ corresponds to a possible outcome of $X$. If it is all (notationally) a bit confusing, this is because the theory was originally developed independently (and not always in the most general way) for discrete events and for real-valued random variables, and the concept of probability space was developed afterwards to create a uniform framework capable of capturing these and more.  --Lambiam 10:39, 11 February 2021 (UTC)