What is a marginal probability distribution
A marginal probability distribution is a probability distribution in which one of the random variables is eliminated from the joint probability distribution. For example, the marginal probability of X is the probability that event X will occur regardless of other events.
For a discrete random variable, the marginal probability distribution is expressed by the following equation:
{\displaystyle P(X)=\sum _{y}P(X,Y)}
For continuous random variables, the marginal probability distribution is expressed by the following equation:
{\displaystyle f_(x)=\int_{y}f(x,y)\,\mathrm {d} y}
As an example of the marginal distribution of a discrete random variable, I will use the following distribution of blood types of boys and girls in one elementary school class.
| X\Y |
Type A |
Type B |
Type O |
Type AB |
| Boy |
0.25 |
0.10 |
0.10 |
0.05 |
| Girl |
0.20 |
0.20 |
0.05 |
0.05 |
The total probability of gender and blood type can then be calculated as follows.
| X\Y |
Type A |
Type B |
Type O |
Type AB |
Sum |
| Boy |
0.25 |
0.10 |
0.10 |
0.05 |
0.50 |
| Girl |
0.20 |
0.20 |
0.05 |
0.05 |
0.50 |
| Sum |
0.45 |
0.30 |
0.15 |
0.10 |
1.00 |
The sum of these values is the marginal probability. Each of these marginal probabilities is shown below.
P(boy)=0.50 \\
P(girl)=0.50 \\
P(type A)=0.45 \\
P(type B)=0.30 \\
P(type O)=0.15 \\
P(type AB)=0.10
Independence of random variables
We can say that X and Y are independent of each other when the random variables X and Y do not affect each other. Independence is determined by whether the joint probabilities can be expressed as a product of the marginal probabilities. The following is an explanation of the discrete and continuous random variable cases, respectively.
Discrete random variable case
We can say that X and Y are independent if the discrete random variables X and Y satisfy the following conditions:
For example, suppose we have the following joint probability distributions for random variables X and Y.
| X\Y |
1 |
2 |
3 |
| 0 |
0.10 |
0.10 |
0.20 |
| 1 |
0.20 |
0 |
0 |
| 2 |
0.10 |
0.10 |
0.20 |
The marginal probabilities are as follows, respectively.
P(X=0)=0.40 \\
P(X=1)=0.20 \\
P(X=2)=0.40 \\
P(Y=1)=0.40 \\
P(Y=2)=0.20 \\
P(Y=3)=0.40
Therefore, P(X=0, Y=1) = 0.10 and P(X=0)P(Y=1) = 0.16 We know that X and Y are not independent because the product of simultaneous and marginal probabilities do not match.
Continuous random variable case
We can say that X and Y are independent if the probability density functions of continuous random variables X and Y satisfy the following:
For example, suppose there is a following probability density function for joint probability:
f(x, y) = \left\{
\begin{array}{ll}
4xy & (0 < x < 1, 0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
For 0 < x < 1, f(x) becomes
\begin{aligned}
f(x) &= \int^{1}_{0} f(x,y) \mathrm{d} y \\
&= \int^{1}_{0} 4xy \mathrm{d} y \\
&= 2x
\end{aligned}
Thus f(x) becomes
f(x) = \left\{
\begin{array}{ll}
2x & (0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
Similarly, f(y) becomes
f(y) = \left\{
\begin{array}{ll}
2y & (0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
Thus, for 0 < x < 1, 0 < y < 1, we can say that x and y are independent since
Next, let us consider the following probability density function for joint probability:
f(x, y) = \left\{
\begin{array}{ll}
x+y & (0 < x < 1, 0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
The f(x) and f(y) are as follows, respectively.
f(x) = \left\{
\begin{array}{ll}
x + \frac{1}{2} & (0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
f(y) = \left\{
\begin{array}{ll}
y + \frac{1}{2} & (0 < y < 1) \\
0 & (otherwise)
\end{array}
\right.
For 0 < x < 1, 0 < y < 1, f(x,y) \neq f(x)f(y), so x and y are not independent.
