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According to Wikipedia, "the expected value of a constant is equal to the constant itself; i.e., if $c$ is a constant, then $\mathbb E[c] = c$." I am currently having a hard time picturing what this means. If I have a random variable $X$ that represents the number of times a coin lands on heads, then I can see how $\mathbb E(X)$ is $0.5$. But $\mathbb E(0.5)$ seems "meaningless." Can someone please explain the significance of the expected value of a constant?

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3 Answers 3

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If it helps your intuition, think of it as a non-random random variable. Something like:

$$ p(y)=\begin{cases}1, &\quad \text{ if } y=c\\ 0, &\quad \text{otherwise} \end{cases} $$

So then the expected value is $$ E(Y)=\sum Pr(Y=y)\times y=c\times1=c. $$

If it's a constant, it can't vary and there's no randomness. So it must be itself, because it cannot be anything else!

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  • $\begingroup$ That clears things up. Thanks. $\endgroup$ Commented Nov 18, 2013 at 11:34
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You are computing the expectation value of the random variable $X$ whose outcome is always the same. Let us focus ourselves to the discrete case, for simplicity. Formally you want to compute the exp. value of the random variable

$$X:(\Omega,P)\rightarrow \operatorname{Im}(X):=\{c\}$$

where $\omega\mapsto X(\omega):=c$ for all $\omega\in\Omega$, with $(\Omega, P)$ finite probability set and $c\in\mathbb R$. Then

$$P_X(X=c):=P(\{\omega\in\Omega~:X(\omega)=c\})=P(\Omega):=1$$

and

$$\mathbb E[X]:=\sum_{c_i\in\operatorname{Im}(X)}c_i\cdot P_X(X=c_i)=c\cdot P_X(X=c)=c\cdot 1=c.$$

The continuous case is similar, with technical differences.

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I think the key here is to view X as it as: a Random Variable.

When writing E(X), x is a constant, you are assuming X to be the random variable of a constant which in turn means its probability value is 1.

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