# Proving Skewness of a Random Variable $X$

• MHB
• mathmari
In summary: Yes, you are correct. I made a typo in my previous calculations. (Oops) In summary, the skewness of a random variable $X$ is defined as $\eta (X):=E\left (\left (\frac{X-\mu }{\sigma}\right )^3\right )$, where $\mu=E(X)$ and $\sigma^2=\text{Var}(X)$. It can also be expressed as $\eta (X)=\frac{E(X^3)-3E(X)E(X^2)+2E(X)^3}{\sigma^3}$. When a random variable is scaled by a positive constant $a$, the skewness remains unchanged, as shown by
mathmari
Gold Member
MHB
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)

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)

## What is skewness?

Skewness is a measure of the asymmetry of a probability distribution. It indicates how much a probability distribution deviates from being symmetrical.

## How is skewness calculated?

Skewness is calculated using the third standardized moment, which is the sum of the cubed deviations from the mean divided by the standard deviation cubed. A positive skewness value indicates a right-skewed distribution, while a negative skewness value indicates a left-skewed distribution.

## How do you prove the skewness of a random variable X?

To prove the skewness of a random variable X, you need to calculate its skewness coefficient using the formula mentioned above. Then, you can compare the calculated skewness value to the critical values for a given level of significance, typically 5%. If the calculated skewness value falls outside of the critical values, you can reject the null hypothesis that the distribution is symmetrical and conclude that the random variable X is skewed.

## What is the null hypothesis in proving skewness of a random variable X?

The null hypothesis in proving skewness of a random variable X is that the distribution is symmetrical, and there is no skewness present. The alternative hypothesis is that the distribution is not symmetrical and has some degree of skewness.

## What are some real-life applications of proving skewness of a random variable X?

Proving skewness of a random variable X is important in many fields, such as finance, economics, and social sciences. It can help identify potential biases in data, assess the risk in financial investments, and understand the distribution of income and wealth in a population. It is also used in quality control to detect any asymmetry in manufacturing processes.

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