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# Calculate variance, standard deviation for conditional and marginal probability distributions

## Variance and Standard Deviation for Marginal Probability Distributions

Generally, the variance for a joint distribution function of random variables $$X$$ and $$Y$$ is given by:

$$Var\left(X,Y\right)=E\left(g\left({x}^2,\ {y}^2\right)\right)-\left(E\left[g\left(x,y\right)\right]\right)^2$$

The standard deviation of joint random variables is the square root of the variance. Therefore, the standard deviation is given by:

$$\sigma_{X,Y}=\sqrt{E\left(g\left({x}^2,\ {y}^2\right)\right)-\left(E\left[g\left(x,y\right)\right]\right)^2}$$

To determine the variance and standard deviation of each random variable that forms part of a multivariate distribution, we first determine their marginal distribution functions and compute the variance and the standard deviation, just like in the univariate case.

### Variance of a Marginal Distribution (Discrete Case)

Suppose a joint distribution of the random variables $$X$$ and $$Y$$ is given by $$P_{XY}(X=x,\ Y=y)$$ (typically abbreviated as $$P_{XY}(x,y)$$) for each pair $$(x,y)$$ of random variables, then for all discrete distributions, two vital requirements hold for each pair $$(x,y)$$:

$$0 \le P_{XY}\left(x,y\right)\le 1$$

And,

$$\sum_{all\ x}\sum_{all\ y}{P_{XY}\left(x,y\right)=1\ }$$

Then, the marginal probabilities $$P_X(X=x)$$ and $$P_Y(Y=y)$$ have variances $$Var(X)$$ and $$Var(Y)$$, respectively:

$$Var\left(X\right)=E\left(X^2\right)-\left[E(X)\right]^2$$

Where

$$E\left(X^2\right)=\sum_{x}{x^2P(X=x)}$$

And

$$E\left(X\right)=\sum_{x}{xP(X=x)}$$

And for the random variable $$Y$$:

$$Var\left(Y\right)=E\left(Y^2\right)-\left[E(Y)\right]^2$$

Where

$$E\left(Y^2\right)=\sum_{x}{y^2P(Y=y)}$$

And

$$E\left(Y\right)=\sum_{x}{yP(Y=y)}$$

### Standard Deviation of a Marginal Distribution (Discrete Case)

The standard deviation is the square root of variance. Denoted by $$\sigma_X$$ and $$\sigma_Y$$, respectively, the variance of $$X$$ and $$Y$$ is given by:

$$\sigma_X=\sqrt{E\left(X^2\right)-\left[E(X)\right]^2}$$

And,

$$\sigma_Y=\sqrt{E\left(Y^2\right)-\left[E(Y)\right]^2}$$

#### Example: Variance and Standard Deviation for Joint Random Variables (Discrete case)

Let $$X$$ and $$Y$$ have joint pmf:

$$f\left(x,y\right)=\frac{x+y}{21},\ \ \ x=1,2,3\ \ \ \ \ y=1,2$$

Calculate the variance and the standard deviation of $$X$$.

Solution

We know that:

$$Var\left(X\right)=E\left(X^2\right)-\left[E(X)\right]^2$$

First, we find the marginal probability mass function of $$X$$, which is given by:

\begin{align*} P_X\left(x\right) & =\sum_{all\ x}{P\left(x,y\right)=}P\left(X=x\right),\ \ \ \ x\epsilon S_x \\ & =\frac{x+\left(1\right)}{21}+\frac{x+\left(2\right)}{21}=\frac{2x+3}{21} \end{align*}

Then,

\begin{align*} E\left(X\right) & =\sum_{\left(all\ x\right)}{xP_X\left(x\right)} \\ & =\sum_{x=1}^{3}{xP_X\left(x\right)=\sum_{x=1}^{3}{x\frac{2x+3}{21}}} \\ & =\left(1\right)\frac{2\left(1\right)+3}{21}+\left(2\right)\frac{2\left(2\right)+3}{21}+\left(3\right)\frac{2\left(3\right)+3}{21}\\ &=1\left(\frac{5}{21}\right)+2\left(\frac{7}{21}\right)+3\left(\frac{9}{21}\right) \\ &=\frac{46}{21} \approx 2.1907619 \end{align*}

And

\begin{align*} E\left(X^2\right) & =\sum_{all\ x}{x^2P_X\left(x\right)} \\ & =\left(1\right)^2\left(\frac{5}{21}\right)+\left(2\right)^2\left(\frac{7}{21}\right)+\left(3\right)^2\left(\frac{9}{21}\right)\\ & =\frac{38}{7} \end{align*}

Thus,

\begin{align*} Var\left(X\right) & =E\left(X^2\right)-\left[E(X)\right]^2 \\ & =\frac{38}{7}-\left(\frac{46}{21}\right)^2=\frac{278}{441}\approx0.6304 \end{align*}

We know that the standard deviation of $$X$$ is the standard deviation of its variance. Therefore,

\begin{align*} \sigma_X & =\sqrt{Va\left(X\right)} \\ & =\sqrt{0.6304}=0.7840 \end{align*}

### Variance of a Marginal Distribution (Continuous case)

We denote the pdf of a joint distribution of the random variables $$X$$ and $$Y$$ by $$f_{XY}(x,y)$$. All continuous distributions must meet two main requirements for each ordered pair $$(x,y)$$ in the domain of $$f$$. The properties include:

$$f_{XY}\left(x,y\right)\geq 0$$

And,

$$\int_{x}\int_{y}{f_{XY}\left(x,y\right)=1}$$

Just like in the discrete case, we can calculate the variances by first calculating their marginal distributions, $$f_X(x)$$ and $$f_Y(y)$$. Recall that:

$$f_X\left(x\right)=\int_{y}{f_{XY}\left(x,y\right)dy}$$

And

$$f_Y\left(y\right)=\int_{x}{f_{XY}\left(x,y\right)dx}$$

Moreover, similar to the discrete case:

$$Var\left(X\right)=E\left(X^2\right)-\left[E(X)\right]^2$$

Where:

$$E\left(X^2\right)=\int_{x}{x^2f_X\left(x\right)dx}$$

And

$$E\left(X\right)=\int_{x}{xf_X\left(x\right)dx}$$

Similarly,

$$Var\left(Y\right)=E\left(Y^2\right)-\left[E(Y)\right]^2$$

Where

$$E\left(Y^2\right)=\int_{y}{y^2f_Y\left(y\right)dy}$$

And

$$E\left(Y\right)=\int_{y}{yf_Y\left(y\right)dy}$$

### Standard Deviation of a Marginal Distribution (Continuous case)

We know that the standard deviation is the square root of the variance:

$$\sigma_X=\sqrt{Var\left(X\right)}$$

And,

$$\sigma_Y=\sqrt{Var\left(Y\right)}$$

#### Example: Variance and Standard Deviation for a Marginal Distribution (Continuous case)

Let $$X$$ and $$Y$$ have the joint pdf:

$$f\left(x,y\right)=\left(\frac{4}{3}\right)\left(1-xy\right),\ \ \ \ \ 0\le x\le1,\ \ \ 0\le y\le1$$

Find $$Var(Y)$$ and $$\sigma_Y$$.

Solution

First, let us find the marginal probability density for $$Y$$.

We know that,

\begin{align*} f_Y\left(y\right) & =\int_{x}{f_{XY}\left(x,y\right)dx} \\ & =\frac{4}{3}\int_{0}^{1}{\left(1-xy\right)\ \ dx} \\ & =\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right),\ \ \ 0\le y\le1\ \\ & =\frac{4}{3}\int_{0}^{1}{\left(1-xy\right)\ \ dx} \\ & =\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right),\ \ \ 0\le y\le1 \end{align*}

Then,

\begin{align*} E\left(Y\right)& =\int_{y}{yf_Y\left(y\right)dy}=\int_{0}^{1}y\left(\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right)\right)dy \\ &=\left[\frac{4}{6}y^2-\frac{4y^3}{18}\right]_0^1 \\ &=\frac{4}{6}-\frac{4}{18} \\ & =\frac{4}{9} \end{align*}

Also,

\begin{align*} E\left(Y^2\right) & =\int_{y}{y^2f_Y\left(y\right)dy}\\ &=\int_{0}^{1}{y^2\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right)\ dy}\\ &=\int_{0}^{1}{y^2\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right)\ dy}\\ &=\int_{0}^{1}{\frac{4y^2}{3}-\frac{4y^3}{6}dy}\\ &=\left|\frac{4y^3}{9}-\frac{4y^4}{24}\right|_0^1=\left(\frac{4}{9}-\frac{4}{24}\right) \\ & =\frac{5}{18} \end{align*}

Therefore,

\begin{align*} & =\int_{0}^{1}{\frac{4y^2}{3}-\frac{4y^3}{6}dy=\left|\frac{4y^3}{9}-\frac{4y^4}{24}\right|_0^1=\left(\frac{4}{9}-\frac{4}{24}\right)=\frac{5}{18}} \\ Var \left(Y\right) & =E\left(Y^2\right)-\left[E\left(Y\right)\right]^2=\frac{5}{18}-\left(\frac{4}{9}\right)^2=\frac{13}{162} \approx 0.080246913 \\ \end{align*}

And now for finding the standard deviation:

\begin{align*} \sigma_Y & =\sqrt{Var\left(Y\right)} \\ &=\sqrt{\frac{13}{162}}=0.2833=0.283278861 \\ \end{align*}

## Variance and Standard Deviation for Conditional Distributions

### Variance of a Conditional Distribution (Discrete Case)

The conditional variance of $$X$$, given that $$Y=y$$, is defined by:

$$Var(X|Y=y)=E\left(X^2|Y=y\right)-\left[E(X|Y=y)\right]^2$$

Where:

$$E\left(X^2|Y=y\right)=\sum_{x}{x^2.g(x|Y=y)}$$

And

$$E\left(X|Y=y\right)=\sum_{x}{x.g(x|Y=y)}$$

Note that this is analogous to the variance of a single random variable.

Similarly, the conditional variance of $$Y$$ given $$X=x$$ is defined by:

$$Var(Y|X=x)=E\left(Y^2|X=x\right)-\left[E(Y|X=x)\right]^2$$

Where:

$$E\left(Y^2|X=x\right)=\sum_{y}{y^2.h(y|X=x)}$$

And

$$E(Y|X=x)=\sum_{x}{y.h(y|X=x)}$$

#### Example: Conditional Variance (Discrete Case)

Let $$X$$ and $$Y$$ have a joint pmf:

$$f\left(x,y\right)=\frac{x+y}{21},\ \ \ x=1,2,3\ \ \ \ \ y=1,2$$

Find $$Var\left(X\middle| Y=1\right)$$.

Solution

First, we have to find the marginal distribution $$f_Y(y)$$ where we know that:

\begin{align*} P_Y\left(y\right) & =\sum_{all\ y}{P\left(x,y\right)=P\left(Y=y\right),\ \ \ \ y\epsilon S_y} \\ {\Rightarrow P}_Y\left(y\right) & =\frac{\left(1\right)+y}{21}+\frac{\left(2\right)+y}{21}+\frac{\left(3\right)+y}{21}=\frac{6+3y}{21} \end{align*}

Secondly, the conditional distribution of $$X$$ given that $$Y=y$$, $$g\left(x\middle| y\right)$$, is equal to:

$$g\left(x\middle| y\right)=\frac{f\left(x,y\right)}{f_Y\left(y\right)}=\frac{\frac{x+y}{21}}{\frac{6+3y}{21}}=\frac{x+y}{3y+6}$$

We know that,

$$Var\left(X\middle| Y=y\right)=\sigma_{X|Y=y}^2\ \ =E\left(X^2|Y=y\right)-\left[E(X|Y=y)\right]^2$$

We need:

$$Var(X|Y=1)=E\left(X^2|Y=1\right)-\left[E(X|Y=1)\right]^2$$

We need to find $$E\left(X\middle| Y=1\right)$$.

\begin{align*} E\left(X\middle| Y=1\right) & =\sum_{x=1}^{3}{x.g\left(x\middle| y=1\right)} \\ & =\sum_{x=1}^{3}{x\frac{\left(x+\left(1\right)\right)}{3\left(1\right)+6}} \\ & =\left(1\right)\frac{\left(1+\left(1\right)\right)}{3\left(1\right)+6}+2\frac{\left(2+\left(1\right)\right)}{3\left(1\right)+6}+3\frac{\left(3+\left(1\right)\right)}{3\left(1\right)+6} \\ &=1\left(\frac{2}{9}\right)+2\left(\frac{1}{3}\right)+3\left(\frac{4}{9}\right) \\ & =\frac{20}{9} \end{align*}

Also, we need,

\begin{align*} E\left(X^2|Y=y\right) & =\sum_{x=1}^{3}{x.g\left(x\middle| y=1\right)} \\ & =\sum_{x=1}^{3}{x^2\frac{\left(x+\left(1\right)\right)}{3\left(1\right)+6}} \\ & =\left(1^2\right)\frac{\left(1+\left(1\right)\right)}{3\left(1\right)+6}+2^2\frac{\left(2+\left(1\right)\right)}{3\left(1\right)+6}+3^2\frac{\left(3+\left(1\right)\right)}{3\left(1\right)+6} \\ & =1\left(\frac{2}{9}\right)+4\left(\frac{1}{3}\right)+9\left(\frac{4}{9}\right) \\ & =\frac{50}{9} \\ \Rightarrow Var\left(X\middle| Y=1\right) & =E\left(X^2|Y=1\right)-\left[E\left(X\middle| Y=1\right)\right]^2 \\ & =\frac{50}{9}-\left(\frac{20}{9}\right)^2=\frac{50}{81} \end{align*}

### Standard Deviation of a Conditional Distribution (Discrete Case)

The conditional standard deviation of $$X$$ given that $$Y=y$$ is defined by:

$$\sigma_{X|Y=y}=\sqrt{Var\left(X\middle| Y=y\right)}$$

While, the conditional standard deviation variance of $$Y$$ given $$X=x$$ is defined by:

$$\sigma_{Y|X=x}=\sqrt{Var\left(Y\middle| X=x\right)}$$

### Variance of a Conditional Distribution (Continuous Case)

The conditional variance of $$X$$ given $$Y=y$$ is:

$$Var\left(X\middle| Y=y\right)=E\left(X^2\middle| Y=y\right)-\left[E\left(X\middle| Y=y\right)\right]^2$$

Where:

$$E\left(X^2\middle| Y=y\right)=\int_{x}{x^2g(x|Y=y)}$$

And

$$E\left(X\middle| Y=y\right)=\int_{x}{x.g(x|Y=y)}$$

Using the same logic,

$$Var\left(Y\middle| X=x\right)=E\left[Y^2\middle| X\right]-\left[E\left(Y\middle| X\right)\right]^2$$

Where:

$$E\left(Y^2\middle| X=x\right)=\int_{y}{y^2h(y|X=x)dy}$$

And

$$E\left(Y\middle| X=x\right)=\int_{y}{y.h(y|X=x)dy}$$

### Standard Deviation of a Conditional Distribution (Continuous Case)

$$\sigma_{X|Y}=\sqrt{Var\left(Y\middle| X=x\right)}$$

And,

$$\sigma_{Y|X}=\sqrt{Var\left(X\middle| Y=y\right)}$$

#### Example: Conditional Variance (Continuous case)

Let the joint pdf of random variables $$X$$ and $$Y$$ be given by:

$$f\left(x,y\right)=\left(\frac{4}{3}\right)\left(1-xy\right),\ \ \ \ \ 0\le x\le1,\ \ \ 0\le y\le1$$

Calculate $$Var\left(X\middle| Y=1\right)$$.

Solution

First, let us find the marginal probability density for $$Y$$.

We know that,

\begin{align*} f_Y\left(y\right) & =\int_{x}{f_{XY}\left(x,y\right)dx} \\ & =\frac{4}{3}\int_{0}^{1}{\left(1-xy\right)\ \ dx} \\ & =\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right),\ \ \ 0\le y\le1\ \end{align*}

We also need we need the conditional distribution $$g(X|Y=y)$$ given by:

$$g\left(X\middle| Y=y\right)=\frac{f_{XY}(x,y)}{f_Y\left(y\right)}=\frac{\left(\frac{4}{3}\right)\left(1-xy\right)}{\left(\frac{4}{3}\right)\left(1-\frac{y}{2}\right)}=\frac{1-xy}{1-\frac{y}{2}}$$

We know that:

$$Var\left(X\middle| Y=y\right)=E\left(X^2\middle| Y=y\right)-\left[E\left(X\middle| Y=y\right)\right]^2$$

But we need:

$$Var\left(X\middle| Y=1\right)=E\left(X^2\middle| Y=1\right)-\left[E\left(X\middle| Y=1\right)\right]^2$$

\begin{align*} E\left(X^2\middle| Y=1\right) & =\int_{x}{x^2.g\left(x\middle| Y=1\right)}dx \\ &=\int_{0}^{1}{x^2.\frac{1-(1)x}{1-\frac{1}{2}}=\int_{0}^{1}{2x^2(1-x)dx=\frac{1}{6}\ }} \end{align*}

Also, we need:

\begin{align*} E\left(X\middle| Y=1\right) & =\int_{x}{x.g\left(x\middle| Y=1\right)}dx \\ &=\int_{0}^{1}{x.\frac{1-(1)x}{1-\frac{1}{2}}=\int_{0}^{1}{2x(1-x)dx=\frac{1}{3}\ }} \\ \Rightarrow Var\left(X\middle| Y=1\right) & =E\left(X^2\middle| Y=1\right)-\left[E\left(X\middle| Y=1\right)\right]^2 \\ &=\frac{1}{6}-\left(\frac{1}{3}\right)^2=\frac{1}{18} \end{align*}

Learning Outcome

Topic 3.e: Multivariate Random Variables – Calculate Variance, the standard deviation for conditional and marginal probability distributions.

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