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I have this statement:

Justify if it is correct or not, perform the following transformation:

Let $X$ a random var, thus:

$P = E(X) - X + 1$

$P = E(1 * X) - X + 1$

$P = X[E(1)] - X + 1$

With the property, $E(c) = c$, then:

$P = X*1 - X +1, => X - X + 1 = 1$

I think the trap step is to get X out of the expected value, but I do not know the explanation of this. Thanks in advance.

ESCM
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  • Going from $E[1 \times X]$ to $X\times E[1]$ is not correct. What is correct is that $E[1 \times X]=E[X]\times E[1]$ and also $E[P]=1$ but neither of these help you – Henry Jul 07 '19 at 22:55

1 Answers1

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This is not correct since the linearity of the expected value (which is just an integral) is just the "constant rule": For any real random variable $X$ with finite expectancy, real constant $c$:

$$\Bbb E(c \cdot X) = \int_\Omega c \cdot X \,\mathrm d \Bbb P= c \int_\Omega X \,\mathrm d \Bbb P = c\cdot \Bbb E(X).$$

Can you see why the above steps fail (in general) if we replace $c$ by another random variable $Y$?

About your particular "riddle": Just consider $X\sim \mathcal N(0,1)$ - a standard normal distribution. Then $\Bbb E(X) - X + 1$ is simply $1-X\sim 1+X\sim \mathcal N(1,1)$, which doesn't equal $0$.

Maximilian Janisch
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  • Therefore is correct that $E(X^2) = X[E(X)]$, but is incorrect that , $E(cX) = X[E(c)] = X*c$? – ESCM Jul 07 '19 at 23:06
  • Neither are correct since $\Bbb E(X^2)$ and $\Bbb E(c X)$ are real numbers, but $c X$ and $\Bbb E(X)X$ are random variables.

    It is correct that $\Bbb E(cX)=c \Bbb E(X)$, but not in general that $\Bbb E(X Y) = \Bbb E(X) \Bbb E (Y)$.

     – Maximilian Janisch Jul 07 '19 at 23:07
  • If $c$ is a constant, and this step is correct $E(cX) = X[E(c)] $, what step is incorrect? or perhaps, when you have $E(Xc)$ can we only remove the constant? – ESCM Jul 07 '19 at 23:11
  • $\Bbb E(cX)\color{red}{\mathbf=}\Bbb E(c)X=cX$ (wrong equality marked in red). You said correctly that "we can only remove the constant". So the following is correct: $\Bbb E(cX)\color{green}=c \Bbb E(X)$. – Maximilian Janisch Jul 07 '19 at 23:12
  • Therefore, when you have $E(Xc)$ can we only remove the constant?, i.e $c[E(X)]$ – ESCM Jul 07 '19 at 23:14
  • @EduardoS. yes! – Maximilian Janisch Jul 07 '19 at 23:14