Proving Skewness of a Random Variable $X$

[mathmari]

Hey!

For a random variable $X$ the skewness is defined by \begin{equation*}\eta (X):=E\left (\left (\frac{X-\mu }{\sigma}\right )^3\right )\end{equation*} where $E(X)=\mu$ and $\text{Var}(X)=\sigma^2$.

I want to show that \begin{equation*}\eta (X)=\frac{E(X^3)-3E(X)E(X^2)+2E(X)^3}{\sigma^3}\end{equation*} and that \begin{equation*}a>0 \Rightarrow \eta (aX+b)=\eta (X)\end{equation*} I have done the following:

Is $\sigma$ a real constant? So by linearity of the expected value we can write the term $\frac{1}{\sigma^3}$ outside of $E$, right? (Wondering)

\begin{align*}\eta (X)&=E\left (\left (\frac{X-\mu }{\sigma}\right )^3\right ) \\ & = E\left ( \frac{(X-\mu)^3 }{\sigma^3} \right ) \\ & = \frac{1}{\sigma^3}\cdot E\left ( (X-\mu)^3 \right )\\ & = \frac{E\left (X^3-3X^2\mu+3X\mu^2-\mu^3 \right )}{\sigma^3} \\ & =\frac{E\left ( X^3-3X^2(E(X))+3X(E(X))^2-(E(X))^3 \right )}{\sigma^3} \\ & = \frac{E[ X^3]-E[3X^2(E(X))]+E[3X(E(X))^2]-E[(E(X))^3]}{\sigma^3} \\ & = \frac{E[ X^3]-3E[X^2(E(X))]+3E[X(E(X))^2]-E[(E(X))^3]}{\sigma^3} \end{align*}

Is everything correct so far? How could we continue? (Wondering)

 

[I like Serena]

Is $\sigma$ a real constant? So by linearity of the expected value we can write the term $\frac{1}{\sigma^3}$ outside of $E$, right?

Is everything correct so far? How could we continue?

Hey mathmari! (Smile)

Yes, $\sigma$ is indeed a constant, so we can move it outside of $E$.
Additionally, $\mu=EX$ is also a constant, so it can be moved outside of $E$ as well.
Moreover, any expectation is a constant. (Thinking)

 

[mathmari]

Yes, $\sigma$ is indeed a constant, so we can move it outside of $E$.
Additionally, $\mu=EX$ is also a constant, so it can be moved outside of $E$ as well.
Moreover, any expectation is a constant. (Thinking)

Ah ok!

So, we get the following?
\begin{align*}\eta (X)& = \frac{E[ X^3]-3E[X^2(E(X))]+3E[X(E(X))^2]-E[(E(X))^3]}{\sigma^3} \\ & = \frac{E[ X^3]-3E(X)E[X^2]+3(E(X))^2E[X]-(E(X))^3E[1]}{\sigma^3}\\ & = \frac{E[ X^3]-3E(X)E[X^2]+3(E(X))^3-(E(X))^3E[1]}{\sigma^3} \end{align*}

Is the last term $E[1]$ correct? If yes, which is its value? (Wondering)

 

[I like Serena]

Ah ok!

Is the last term $E[1]$ correct? If yes, which is its value?

An expectation is what we expect some random variable to be on average.
It the only possible outcome is a constant, the expectation must be that same constant on average, mustn't it? (Wondering)
In other words, there is no need to introduce $E[1]$. We just have that $E[(EX)^3]=(EX)^3$.

 

[mathmari]

An expectation is what we expect some random variable to be on average.
It the only possible outcome is a constant, the expectation must be that same constant on average, mustn't it? (Wondering)
In other words, there is no need to introduce $E[1]$. We just have that $E[(EX)^3]=(EX)^3$.

I see! (Nerd) About the second part:

We have that \begin{align*}\eta (aX+b)&=\frac{E[ (aX+b)^3]-3E(aX+b)E[(aX+b)^2]+2[E(aX+b)]^3}{\sigma^3}\\ & =\frac{E[ a^3X^3+3a^2X^2b+3aXb^2+b^3]-3E(aX+b)E[a^2X^2+2aXb+b^2]+2[a^3(E(X))^3+3a^2(E(X))^2b+3aE(X)b^2+b^3]}{\sigma^3}\\ & = \frac{ a^3E[X^3]+3a^2bE[X^2]+3ab^2E[X]+b^3-3\left (aE[X]+b\right )\left (a^2E[X^2]+2abE[X]+b^2\right )+2[a^3(E(X))^3+3a^2(E(X))^2b+3aE(X)b^2+b^3]}{\sigma^3} \\ & = \frac{ a^3E[X^3]+3a^2bE[X^2]+3ab^2E[X]+b^3-3[a^3E[X^2]E[X]+2a^2b(E[X])^2+ab^2E[X]+a^2bE[X^2]+2ab^2E[X]+b^3 ]+2[a^3(E(X))^3+3a^2(E(X))^2b+3aE(X)b^2+b^3]}{\sigma^3} \\ & =\frac{ a^3E[X^3]+3a^2bE[X^2]+3ab^2E[X]+b^3-3a^3E[X^2]E[X]-6a^2b(E[X])^2-3ab^2E[X]-3a^2bE[X^2]-6ab^2E[X]-3b^3 +2a^3(E(X))^3+6a^2(E(X))^2b+6aE(X)b^2+2b^3}{\sigma^3} \\ & = \frac{ a^3E[X^3]-3a^3E[X^2]E[X] +2a^3(E(X))^3}{\sigma^3} \\ & = a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{\sigma^3}\\ & = a^3\cdot \eta (X) \end{align*}

Is everything correct?

But at the exercise it is $a>0\Rightarrow \eta (aX+b)=\eta (X)$. Is it a typo or I have I done something wrong? (Wondering)

 

[I like Serena]

Is everything correct?

But at the exercise it is $a>0\Rightarrow \eta (aX+b)=\eta (X)$. Is it a typo or I have I done something wrong?

You're assuming that $\sigma$ is the same, but I'm afraid it isn't. (Thinking)

 

[mathmari]

You're assuming that $\sigma$ is the same, but I'm afraid it isn't. (Thinking)

Ah ok.

We have that $$\sigma=\text{Var}(X)=\mathbb{E}[X^2]-\left (\mathbb{E}[X]\right )^2$$ So, when we have $aX+b$ then we get \begin{align*}\tilde{\sigma}&= \text{Var}(aX+b)\\ & =E[(aX+b)^2]-\left (E[aX+b]\right )^2\\ & =E[a^2X^2+2abX+b^2]-\left (aE[X]+b\right )^2\\ & =a^2E[X^2]+2abE[X]+b^2-\left (a^2(E[X])^2+2abE[X]+b^2\right ) \\ & = a^2E[X^2]+2abE[X]+b^2-a^2(E[X])^2-2abE[X]-b^2 \\ & = a^2E[X^2]-a^2(E[X])^2\\ & = a^2\cdot \left (E[X^2]-(E[X])^2\right ) \\ & =a^2\cdot \sigma \end{align*}
right?

Then we would get \begin{align*}\eta (aX+b)&=a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{\tilde\sigma^3} \\ & =a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{(a^2\cdot \sigma )^3}\\ & = a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{a^6\cdot \sigma ^3}\\ & = \frac{1}{a^3}\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{ \sigma ^3} \\ & = \frac{1}{a^3}\cdot \eta (X)
\end{align*}

Where is my mistake? (Wondering)

 

[I like Serena]

We have that $$\sigma=\text{Var}(X)=\mathbb{E}[X^2]-\left (\mathbb{E}[X]\right )^2$$

Isn't it
$$\sigma^{\color{red}2}=\text{Var}(X)=\mathbb{E}[X^2]-\left (\mathbb{E}[X]\right )^2$$
(Wondering)

 

[mathmari]

Isn't it
$$\sigma^{\color{red}2}=\text{Var}(X)=\mathbb{E}[X^2]-\left (\mathbb{E}[X]\right )^2$$
(Wondering)

Oh yes (Tmi)

Then we have that $\tilde\sigma^2=a^2\cdot \sigma^2$.

Therefore get \begin{align*}\eta (aX+b)&=a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{\tilde\sigma^3} \\ & = a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{(\tilde\sigma^2)^\frac{3}{2}} \\ & =a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{(a^2\cdot \sigma^2 )^{\frac{3}{2}}}\\ & = a^3\cdot \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{a^3\cdot \sigma ^3}\\ & = \frac{ E[X^3]-3E[X^2]E[X] +2(E(X))^3}{ \sigma ^3} \\ & = \eta (X)
\end{align*} (Whew)