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I am asked to find the expected value of a vector of two random variables when the joint density is given. Is the recipe for solving this problem:

  • Find the marginal distributions
  • Find the expected values of the marginal distributions

which will involve a relatively long integration process, since I have to find the two marginals and then the two expected values?

Or is there some shortcut?

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    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$
    – gung - Reinstate Monica
    Dec 8, 2015 at 17:12
  • $\begingroup$ @Xi'an What are the double integrals I should solve? Something like $\int\int xyf(x,y)dxdy$? $\endgroup$
    – vec
    Dec 8, 2015 at 17:15
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    $\begingroup$ @gung yes; tag added $\endgroup$
    – vec
    Dec 8, 2015 at 17:16
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    $\begingroup$ The expected value of a vector $(X,Y)$ random variable is defined to be $E(X,Y])= (E[X], E[Y])$, that is, the vector of the individual expectations. So you don't really have a "shortcut" except that you can hide what you are doing in a double integral. $$E[X] = \int_{-\infty}^\infty x\cdot f_X(x)\,\mathrm dx = \int_{-\infty}^\infty x \left[\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy\right]\,\mathrm dx = \int_{-\infty}^\infty \int_{-\infty}^\infty x\cdot f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx,$$ and similarly for $Y$. Is solving the rightmost integral the same as first finding the marginal? $\endgroup$
    – Dilip Sarwate
    Dec 8, 2015 at 18:08

2 Answers 2

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As rightly pointed out by Dilip Sarwate, computing the expectation of one component as a two dimensional integral requires integrating out the other element of the vector: $$\mathbb{E}[X] = \iint x\,f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx,$$ (which is a special case of the so-called law of the uncounscious statistician). The only simplifications I can think of is

  1. when finding the conditional expectation of one component given the other is easier: $$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|Y]] = \int x\,f_{X|Y}(x|y)\,\mathrm dx\,f_Y(y)\,\mathrm dy,$$ by the double projection theorem, in the sense that this could require computing a single marginal instead of two;
  2. when computing the marginal cdf is easier, since $$\mathbb{E}[X] = \int_{-\infty}^0 F_X(x)\,\mathrm dx-\int_{+\infty}^0 (1-F_X(x))\,\mathrm dx$$ by an integration by parts.
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  • $\begingroup$ In #1, about the conditional, are there supposed to be two $\mathrm dx$ in the expression? Forgive my ignorance, I'm not sure how to understand the expression as it is right now given there's only one integral symbol and two $\mathrm dx$. The link to the reference is broken, and Archive.org's Wayback Machine doesn't show the LaTeX equations. $\endgroup$
    – Ahmed Fasih
    Aug 25, 2021 at 23:18
  • $\begingroup$ @AhmedFasih: typo corrected $\endgroup$
    – Xi'an
    Aug 26, 2021 at 5:22
  • $\begingroup$ FYI, 1. is called "The Law of Iterated Expectation" or "Adam's law". $\endgroup$
    – Abhishek Divekar
    Aug 19 at 15:24
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Using Monte Carlo estimate, the expected value for the marginal distribution will the $\frac{\displaystyle\sum_{i=1}^Nx_i}{N}$ where $x_i$ are x's samples from the (x,y) samples from joint distribution.

By Monte Carlo integration, $\int\int\phi(x,y)f(x,y)dxdy$ can be estimated by $\frac{1}{N}\displaystyle\sum_{i=1}^N\phi(x_i,y_i)$. Here, we have $\phi(x,y)= x$.

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    $\begingroup$ Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. $\endgroup$
    – gung - Reinstate Monica
    Dec 5, 2017 at 0:10

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