2
$\begingroup$

In secondary education, random variables aren't rigorously defined. Contrary to tertiary education, where they are usually maps $X\colon \;\Omega \rightarrow \mathbb{R}$ with $\Omega$ the sample space of the probability space, they are just seen as placeholders for some undetermined real quantity which assumes certain values with a well-defined probability.

If random variables are defined rigorously, the proof of the linearity of the expectation value is trivial, it follows (with $a,b$ real) from the linearity of the integral (or of summation):

$$\mathbf{E}[a X+ b Y] = \sum_{\omega \in \Omega} \mathbf{P}[\omega]\cdot\bigl(a X(\omega)+ b Y(\omega)\bigr) = \\ = a\sum_{\omega \in \Omega} \mathbf{P}[\omega]\cdot X(\omega) + b\sum_{\omega \in \Omega} \mathbf{P}[\omega] \cdot Y(\omega) = a\,\mathbf{E}[X] + b\,\mathbf{E}[Y]\,. $$

But what would be the best way to prove/explain, that the expectation value is linear if one hasn't defined random variables rigorously? After all, in the non-rigorous "definition", $X$ and $Y$ are just known by their possibly different probability distributions on the same sample space, $\mathbb{R}$.

  • 0
    A somewhat similar question was asked two days ago and is linked at the end of this comment. The accepted answer from that question at least gives a simple example illustrating the linearity of expectation. http://math.stackexchange.com/questions/2080030/linearity-of-expectations-why-does-it-hold-intuitively-even-when-the-r-v-s-are – 2017-01-04

1 Answers 1

1

I guess in the non-rigorous approach you'd still have definitions like $$ E(X) = \int xP(x,y,z) dxdydz $$ and $$ E(Y) = \int y P(x,y,z)dxdydz $$

for, say, a setting with continuous RVs X,Y,Z. So $E(aX+bY) = aE(X) + bE(Y)$ would still follow from the linearity of these integrals. (Or sums in the discrete case).

We can also get to the formula $$ E(aX+bY) = \int (ax+by)P(x,y)dxdy$$ starting from $Z = aX+bY$ and the formula for the expectation value for $Z$ $$E(aX+bY) = \int zP_Z(z) dz.$$

Here, we need to use the formula for the PDF of the sum of two RVs, $P_Z(z) = \int P_{aX,bY}(\xi, z-\xi)d\xi$ where $P_{aX,bY}$ is the joint PDF for $aX$ and $bY.$ Then we transform from $(aX,bY)$ to $(X,Y)$ and get $P_{X,Y}(x,y) = abP_{aX,bY}(ax,by)$ so that $$ P_Z(z) = \int \frac{1}{ab} P_{X,Y}\left(\frac{\xi}{a},\frac{z-\xi}{b}\right)d\zeta.$$

Finally, plug this into the formula for expectation and get $$ E(aX+bY = \int zdz\int d\xi \frac{1}{ab}P_{X,Y}\left(\frac{\xi}{a},\frac{z-\xi}{b}\right)= \int zdz\int dx \frac{1}{b}P_{X,Y}\left(x,\frac{z-ax}{b}\right). $$ Where we substituted $x = \xi/a.$ Now, change the order of integration and sub $y = (z-ax)/b$ for $z$ to get $$ E(aX+bY) = \int dx \int zdz\frac{1}{b}P_{X,Y}\left(\xi,\frac{z-a\xi}{b}\right)= \int dx \int dy (ax+by)P_{X,Y}(x,y)$$ and again, the linearity of the integral shows this is equal to $aE(X) + bE(Y).$

  • 0
    Could you give a full "proof"? I think the problem is, that naively one starts with the combined random variable $aX+bY$, so that we must get from $\mathbf{E}[aX+bY]=\sum_z z\cdot \mathbf{P}[aX+bY=z] = \dots$ to $\dots = a\mathbf{E}[X] + b\mathbf{E}[Y]$. – 2017-01-08
  • 0
    @wolf-revo-cats Ok, I edited to derive that. It's for continuous RVs (the discrete is a bit more of a headache). The more natural way of looking at $E(aX+bY)$ is as a sum or integral over the joint distribution of $X$ and $Y$ rather than over the distribution of $Z=aX+bY.$ (In fact, I think my 'derivation' is better understood backwards as a derivation of the formula for the PDF of the sum of two RVs.) – 2017-01-08