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Distribution of random variable

Probability distribution 

Random variables have distributions. They can have many forms, however each practice has the one that serves it the best.  For better understanding, we can visualize distribution as a tree diagram of a random variable with a little bit of a changed shape.

Probability distribution of discrete random variable is a list of probabilities associated with each of its possible values.

Example. Lets toss two coins 4 times. Let $X$ be the number of tails.

$\Omega=$set of all possible outcomes$=\{HH, HT, TH, TT\}$. The probability of each outcome $\omega \in \Omega$ is $P(\omega)=\frac{1}{4}$. We have 3 possibilities:

1. we got 0 tails
2. we got 1 tail
3. we got 2 tails

Lets calculate the probability of each one:

      1. The only outcome where we got no tails is $\{HH\}$ so let’s calculate its probability $$P(X=0)=P(\{HH\})=\frac{1}{4}$$
      2. Now for 1 tail, we have two outcomes $\{HT, TH\}$ $$P(X=1)=P(\{HT, TH\})=\frac{1}{2}$$
      3. For 2 tails, there is only one outcome $\{TT\}$ $$P(X=2)=P(\{TT\})=\frac{1}{4}$$

Consequently, the table of distribution would look like this :

In the top row we put the outcomes we were looking for. In the bottom row we put the probability of each outcome.

Generally,  for $X(\Omega)=\{a_{1}, a_{2}, …, a_{n}, … \}$ and $P(a_{j})=p_{j}$, for $j=1, 2, 3….$ probability distribution looks like this:

Another way of representing probability distribution is by using graphs. The most popular option is probability histogram, but we can use whichever we want. The histogram of previous example would look like this:

The $x$-axis is value for random variable $X$, while the $y$-axis is the probability of each given value.

Other important part of random variable is its mean, which is a measure of central location, and the variance and standard deviation, which are measures of spread.

Example. Random variable $X$ has distribution

for $c\in \mathbb{R}$.

(a) Find $c$.
(b) Calculate the probability of $X$ assuming value between 2 and 5 (both included).
(c) Find the smallest $k\in \mathbb{N}$ that $P(X \leq k) \geq \frac{2}{5}$.

(a) Since $\sum_{i \in \mathbb{N}} p_{i}=1$ we have:

$c+2c+2c+3c+c^{2}+2c^{2}+(7c^{2}+c)=1$

$10c^{2}+9c-1=0$

When we solve this quadratic equation, we get $c_{1}=-1$ and $c_{2}=\frac{1}{10}$. However, since probability can’t be negative, the only solution is $c_{2}=\frac{1}{10}$. As a result, the distribution of $X$ looks like this:

(b) $P(2 \leq X \leq 5)=P(\{2, 3, 4, 5\})$ since all these outcomes are mutually exclusive the expression is equal to

$= P(X=2)+P(X=3)+P(X=4)+P(X=5)= $
$\displaystyle{=\frac{2}{10} + \frac{2}{10} + \frac{3}{10} + \frac{1}{100} = \frac{71}{100}}$

(c) Firstly, lets try with $k=1$.

$\displaystyle{P(X \leq 1)=P(X=1)=\frac{1}{10}}$

However, that is not greater or equal to $\displaystyle{\frac{2}{5}}$.

Secondly, for $k=2$

$\displaystyle{P(X \leq 2)=P(X=1)+P(X=2)=\frac{1}{10} + \frac{2}{10}=\frac{3}{10}}.$

Once again, the result isn’t greater or equal to $\displaystyle{\frac{2}{5}}$.

For $k=3$ we have:

$\displaystyle{P(X \leq 3)=P(X=1)+P(X=2)+P(X=3)=\frac{1}{10} + \frac{2}{10} + \frac{2}{10}= \frac{5}{10}=\frac{1}{2} > \frac{2}{5}}$

As a result, $k=3$.

Probability function 

The another way of representing the probability distribution is by using a function that gives the probability that a discrete random variable is exactly equal to some value. That function is called probability function or probability mass function and noted with PMF.

For a discrete random variable $X$ it is defined as $f_{X}(x)=P(X=x)$.

Example
Suppose $\Omega$ is the sample space of all outcomes of a single toss of a fair coin, and $X$ is the random variable defined on $\Omega$. Assigning 0 to the category “tails” and 1 to the category “heads”. Since the coin is fair, consequently the probability mass function is $$f_{X}(x)=\begin{cases} \frac{1}{2}, & \text{x $\in$ \{0,1\}}.\\ 0, & \text{otherwise}.\end{cases}$$

Similarly for rolling a fair die. $$f_{X}(x)=\begin{cases} \frac{1}{6}, & \text{x $\in$ \{1,2,3,4,5,6\}}.\\ 0, & \text{otherwise}.\end{cases}$$

Note that $\sum_{X\in \Omega} f_{X}(x)=1$

However, PMF is a little bit more complicated for the continuous random variable but that will be explained in another lesson.