Understanding Random variable- a variable takes value with uncertainty.

Lecture note -3

Discrete Random variable: II

In my previous lecture notes on “Random Variable”, the notions of a random variable and discrete random variables have been mentioned. Today we will look into few characteristics of discrete random variables.

Let’s start with the example of lecture note 2, we have considered a random experiment of tossing a biased coin thrice, where P(H)=1/3 and we made a random variable X, that stands for the number of heads obtained in that experiment.

Possible values of X, i.e. support of X is {0, 1, 2, 3} and we have obtained probabilities for each value as follows;

P(X= 0) = 8/27

P(X=1) = 4/9

P(X=2) = 2/9

P(X = 3) =1/27

Here, (because the total probability must be 1). We can think of as we are distributing the total probability of 1 over the values of X using the function “P()”. The function “P()” specifies the values lie between 0 to 1 for each possible values of X in such a manner that if we take the sum of the values specified by P(), then the total will be equal to 1. When such a probability function of any discrete random variable exists then we call it the probability mass function (pmf) of that discrete random variable.

Thus the main purpose of the pmf is to distribute total probability 1 over the possible values of the discrete random variable.

Suppose Y be a discrete random variable, support of Y is {1,2}. Thus Y can take only two values, 1 or 2.

If someone specifies the pmf of Y as, P(Y=1)=1/3 and P(Y=2)=1/2. Is she right?

No! because, P(Y=1) + P(Y=2) ≠ 1 , here the mentioned pmf is not distributing total probability 1 over the possible values of Y.

In the last lecture note, I have asked a question, I hope you all forgot it. I am mentioning here again,

Exercise: Suppose you will be tossing this biased coin thrice, where P(H)=1/3. How many heads are you expecting?

To answer this question we need to know the method of computation of expectation of a discrete random variable.

Here X is the random variable. The notation of the expectation is E(X) which is formulated as follows;

[The basic rule of calculation of expectation for a discrete random variable is very simple, first take the product of the value of the random variable with the probability of that value, then take the sum of the products.]

E(X)= sum of {value of the random variable × Probability of that value}= = 1 (check the calculation).

Thus we are expecting 1 head from this random experiment.

Note: The notion or concept of expectation of a random variable is not very clear from the formal definition or the formula to calculate it, mentioned above. I hope things will be clear if you follow my lecture videos. I will be uploading my lecture video on expectation soon.

Now suppose Y be another discrete random variable, the probability distribution on Y is mentioned by its pmf as follows;

P(Y=i) = 1/10 for all i = 5(1)14

What is the expectation of Y?

Answer with a clarification:

First of all what is that 5(1)14?

It means a sequence of integers starts from 5, increases by 1, and ends at 14. Basically it is 5,6,7, . . . ,13,14.

So, the support of Y is {5, 6, . . . , 13 , 14}. We have 10 possible values (5 to 14) of Y, each of them has probability 1/10 (lies between 0 to 1), thus the total probability, . (If you still have any doubt then you can check thoroughly). Here we have no problem with the mentioned pmf.

Calculation of expectation:

E(Y) = = 5×1/10 + 6×1/10 + ….. + 13×1/10+ 14×1/10 = 9.5 (Answer).

In next my lecture note, I will be discussing Continuous Random Variable.

Give a test on discrete random variable using the following link;

testmoz.com/4899688

Go through the link mentioned above, there is a place where you have to mention your name. Do mention your full name before taking the test.