MHB Proving Skewness of a Random Variable $X$

AI Thread Summary
The discussion focuses on proving the skewness of a random variable $X$, defined as $\eta(X) = E\left(\left(\frac{X - \mu}{\sigma}\right)^3\right)$. Participants confirm that $\sigma$ and $\mu$ can be treated as constants, allowing them to be moved outside the expectation operator. The calculation progresses to show that the skewness can be expressed in terms of the third and second moments of $X$. A key point of confusion arises regarding the transformation of the random variable $aX + b$, where it is clarified that the variance changes, leading to a correction in the skewness formula. Ultimately, the correct conclusion is reached that $\eta(aX + b) = \eta(X)$ for $a > 0$.
mathmari
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Hey! :o

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)
 
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mathmari said:
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)
 
I like Serena said:
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)
 
mathmari said:
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 $E[(EX)^3]=(EX)^3$.
 
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
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)
 
mathmari said:
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)
 
I like Serena said:
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)
 

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