If you are not familiar with the concept of variance, please consult our guide on variance first.
Just as we did in the section of expected values, we will prove the properties of variance for the case of discrete random variables. These properties also hold for the continuous random variables, and their proofs are analogous to the discrete case.
Theorem.
The variance of a random variable $X$ can also be computed by:
$$\mathbb{V}(X)=\mathbb{E}(X^2)-\big[\mathbb{E}(X)\big]^2$$
We typically use this equation instead of the definition to compute the variance of $X$.
Proof. We begin with the definition of variance:
$$\begin{align*}
\mathbb{V}(X)&=\mathbb{E}\big[(X-\mu)^2\big]\\
&=\sum_x(x-\mu)^2\cdot{p(x)}\\
&=\sum_x(x^2-2\mu{x}+\mu^2)\cdot{p(x)}\\
&=\sum_x\Big(x^2\cdot{p(x)}-2\mu{x}\cdot{p(x)}+\mu^2\cdot{p(x)}\Big)\\
&=\sum_x\Big(x^2\cdot{p(x)}\Big)-2\mu\sum_x\Big(x\cdot{p(x)}\Big)+\mu^2\sum_xp(x)\\
&=\mathbb{E}(X^2)-2\mu\cdot\mathbb{E}(X)+\mu^2\\
&=\mathbb{E}(X^2)-2\mu^2+\mu^2\\
&=\mathbb{E}(X^2)-\mu^2\\
&=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2
\end{align*}$$
This completes the proof.
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Theorem.
Variance of aX+b where a and b are constants
Let $X$ be a random variable. If $a$ and $b$ are constants, then:
$$\mathbb{V}(aX+b)=a^2\cdot{\mathbb{V}(X)}$$
Proof. To compute the variance of $aX+b$, we first need to compute the mean value of $aX+b$. To avoid confusion, we will denote the mean of $X$ as $\mu_X$ and the mean of $aX+b$ as $\mu_{aX+b}$. The mean of $aX+b$ is as follows:
$$\begin{align*}
\mu_{aX+b}&=\mathbb{E}(aX+b)\\
&=a\cdot{\mathbb{E}(X)}+b\\
&=a\cdot{\mu_X}+b\\
\end{align*}$$
Now, use the definition of variance:
$$\begin{align*}
\mathbb{V}(aX+b)&=
\mathbb{E}\big[(aX+b-\mu_{aX+b})^2\big]\\
&=\mathbb{E}\big[(aX+b-(a\mu_X+b))^2\big]\\
&=\mathbb{E}\big[(aX-a\mu_X)^2\big]\\
&=\mathbb{E}\big[(a(X-\mu_X))^2\big]\\
&=\mathbb{E}\big[a^2(X-\mu_X)^2\big]\\
&=a^2\cdot\mathbb{E}[(X-\mu_X)^2\big]\\
&=a^2\cdot\mathbb{V}(X)\\
\end{align*}$$
This completes the proof.
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Theorem.
Variance of sum of two random variables
The variance of the sum of two random variables $X$ and $Y$ is:
$$\begin{align*}
\mathbb{V}(X+Y)=\mathbb{V}(X)+\mathbb{V}(Y)+2\cdot{\mathrm{cov}(X,Y)}\\
\end{align*}$$
Where $\text{cov}(X,Y)$ is the covariance of $X$ and $Y$.
Proof. Let the mean of random variables $X$ and $Y$ be $\mu_X$ and $\mu_Y$, respectively. We also denote the mean of $X+Y$ as $\mu_{X+Y}$. From the definition of variance:
$$\begin{align*}
\mathbb{V}(X+Y)
&=\mathbb{E}\big[(X+Y-\mu_{X+Y})^2\big]\\
&=\mathbb{E}\big[(X+Y-\mathbb{E}(X+Y))^2\big]\\
&=\mathbb{E}\big[(X+Y-\mathbb{E}(X)-\mathbb{E}(Y))^2\big]\\
&=\mathbb{E}\big[(X+Y-\mu_X-\mu_Y)^2\big]\\
&=\mathbb{E}\big[\big((X-\mu_X)+(Y-\mu_Y)\big)^2\big]\\
&=\mathbb{E}\big[(X-\mu_X)^2+(Y-\mu_Y)^2+2(X-\mu_X)(Y-\mu_Y)\big]\\
&=\mathbb{E}\big[(X-\mu_X)^2\big]
+\mathbb{E}\big[(Y-\mu_Y)^2\big]
+2\mathbb{E}\big[(X-\mu_X)(Y-\mu_Y)\big]\\
&=\mathbb{V}(X)+\mathbb{V}(Y)+2\cdot\mathrm{cov}(X,Y)
\end{align*}$$
This completes the proof.
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Theorem.
Variance of the difference of two random variables
The variance of the difference of two random variables $X$ and $Y$ is:
$$\mathbb{V}(X-Y)=\mathbb{V}(X)+\mathbb{V}(Y)-2\cdot{\mathrm{cov}(X,Y)}\\$$
Where $\text{cov}(X,Y)$ is the covariance of $X$ and $Y$.
Proof. We know from theoremlink that:
$$\begin{align*}
\mathbb{V}(X+Y)=\mathbb{V}(X)+\mathbb{V}(Y)+2\cdot{\mathrm{cov}(X,Y)}\\
\end{align*}$$
Setting $Y=-Y'$ gives:
$$\begin{align*}
\mathbb{V}(X-Y')
&=\mathbb{V}(X)+\mathbb{V}(-Y')+2\cdot{\mathrm{cov}(X,-Y')}\\
&=\mathbb{V}(X)+\mathbb{V}(Y')-2\cdot{\mathrm{cov}(X,Y')}\\
\end{align*}$$
For the second equality, we used theoremlink and theorem TODO.
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Theorem.
Variance of sum of two independent variables
The variance of a sum of two independent random variables $X$ and $Y$ is:
$$\mathbb{V}(X+Y)=\mathbb{V}(X)+\mathbb{V}(Y)$$
Proof. We know from theoremlink that:
$$\begin{align*}
\mathbb{V}(X+Y)
=\mathbb{V}(X)+\mathbb{V}(Y)+2\cdot{\mathrm{cov}(X,Y)}\\
\end{align*}$$
If $X$ and $Y$ are independent, then their covariance is zero. This completes the proof.
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Theorem.
Interchanging the summation sign and variance
If $X_1,X_2,\cdots,X_n$ are independent random variables, then:
$$\sum_{i=1}^{n}\mathbb{V}(X_i)
=
\mathbb{V}\Big(\sum_{i=1}^{n}X_i\Big)$$
Note that this is only true when the random variables are independent.
Proof. Let's start from the left-hand side:
$$\begin{equation}\label{eq:xHVLlnGdf6OA8hdl19R}
\sum^n_{i=1}\mathbb{V}(X_i)
=\mathbb{V}(X_1)+\mathbb{V}(X_2)+\cdots+\mathbb{V}(X_n)
\end{equation}$$
From theoremlink, we know that $\mathbb{V}(X)+\mathbb{V}(Y)=\mathbb{V}(X+Y)$ if $X$ and $Y$ are independent random variables. Since $X_1,X_2,\cdots,X_n$ are all independent, then \eqref{eq:xHVLlnGdf6OA8hdl19R} becomes:
$$\begin{align*}
\sum^n_{i=1}\mathbb{V}(X_i)
&=\mathbb{V}(X_1+X_2+\cdots+X_n)\\
&=\mathbb{V}\Big(\sum^n_{i=1}X_i\Big)
\end{align*}$$
This completes the proof.
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Theorem.
Law of total variance
The law of total variance states that:
$$\mathbb{V}(X)
=\mathbb{E}\big[\mathbb{V}(X|Y)\big]
+\mathbb{V}\big[\mathbb{E}(X|Y)\big]$$
Here, $\mathbb{E}\big[\mathbb{V}(X|Y)\big]$ is referred to as the within-group variation, while $\mathbb{V}\big[\mathbb{E}(X|Y)\big]$ is referred to as the between-group variation.
Proof. Let us color-code the law of total variance:
$$\begin{equation}\label{eq:hxeXJ27Ezx8yJHNtkdd}
\mathbb{V}(X)
=\mathbb{E}
\big[{\color{green}\mathbb{V}(X|Y)}\big]+
{\color{red}\mathbb{V}\big[\mathbb{E}(X|Y)\big]}
\end{equation}$$
Let's start with the green term. From theoremlink, we have that:
$${\color{green}\mathbb{V}(X|Y)}=
\mathbb{E}(X^2|Y)-\big[\mathbb{E}(X|Y)\big]^2\\$$
Taking the expected value on both sides:
$$\mathbb{E}\big[{\color{green}\mathbb{V}(X|Y)}\big]
={\color{purple}\mathbb{E}\big[\mathbb{E}(X^2|Y)\big]}
-\mathbb{E}\big[\mathbb{E}(X|Y)^2\big]$$
From the law of total expectation TODO, we know that ${\color{purple}\mathbb{E}\big[\mathbb{E}(X^2|Y)\big]}
=\mathbb{E}(X^2)$:
$$\begin{equation}\label{eq:Keng91E84H2xMJ4cEes}
\mathbb{E}\big[{\color{green}\mathbb{V}(X|Y)}\big]=
\mathbb{E}(X^2)-\mathbb{E}\big[\mathbb{E}(X|Y)^2\big]
\end{equation}$$
We then move on to the red term in \eqref{eq:hxeXJ27Ezx8yJHNtkdd}. From theoremlink, we have that:
$${\color{red}\mathbb{V}\big[\mathbb{E}(X|Y)\big]}
=\mathbb{E}\big[\big(\mathbb{E}(X|Y)\big)^2\big]-
{\color{orange}\big[\mathbb{E}\big(\mathbb{E}(X|Y)\big)\big]^2}$$
Again, from the law of total expectation TODO, we have that ${\color{orange}\big[\mathbb{E}\big(\mathbb{E}(X|Y)\big)\big]^2}
=
\big[\mathbb{E}(X)\big]^2$:
$$\begin{equation}\label{eq:GNQl9RdOFbD0Abu3OG6}
{\color{red}\mathbb{V}\big[\mathbb{E}(X|Y)\big]}=
\mathbb{E}\big[\big(\mathbb{E}(X|Y)\big)^2\big]-\big[\mathbb{E}(X)\big]^2
\end{equation}$$
Taking the sum of \eqref{eq:Keng91E84H2xMJ4cEes} and \eqref{eq:GNQl9RdOFbD0Abu3OG6} yields:
$$\begin{align*}
\mathbb{E}\big[{\color{green}\mathbb{V}(X|Y)}\big]
+{\color{red}\mathbb{V}\big[\mathbb{E}(X|Y)\big]}
&=\mathbb{E}(X^2)-\mathbb{E}\big[\mathbb{E}(X|Y)^2\big]+\mathbb{E}\big[\big(\mathbb{E}(X|Y)\big)^2\big]-\big[\mathbb{E}(X)\big]^2\\
&=\mathbb{E}(X^2)-\big[\mathbb{E}(X)\big]^2\\
&=\mathbb{V}(X)
\end{align*}$$
This completes the proof.
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