A number can either be 0 or 1. P(0)=0.5 P(1)=0.5 What is its Variance?

Question

How can be derive variance from probability?

Answer 1

Hi Soham,

 

 

First make a table listing the 2 outcomes (0 and 1) and the probability of each outcome:

 

x    P(x)

=   =

0   1/2

1   1/2

 

You can think of this as flipping a coin once. You can call "success" either getting a head or a tail.

If you call success getting a "head" then "x = 0" means getting zero heads or 1 tail. The probability of that

is 1/2.  "x = 1" means getting 1 head. The probability of that is also 1/2.

 

This may sound very complicated math for a simple problem, but the math is useful for more complicated

cases like flipping a coin multiple times.

 

"x" is called a random variable which just lists the 2 possible outcomes.

 

I am writing the probabilities as fractions because the calculations are easier using fractions.

 

This editor doesn't have an equation editor so it is hard to write formulas so excuse the possibly

confusing notation that follows.

 

Please look up "variance for a probability distribution" to get the formula online.
I will use 1 of the 2 versions of the formula.

 

variance = sigma [ (x - mean)² * P(x) ]

 

where

sigma is the summation of the 2 values of  [ (x - mean)² * P(x) ] for the 2 different x values.

 

the mean (also called the "average" or "arithmetic average" to be precise) is:

 

mean = sigma [ (x * P(x) ]

 

Again, sigma is the sum of the 2 values of x * P(x) for the 2 different x values.

 

 

Let's commute the mean:

 

mean = (0 * 1/2) + (1* 1/2) = 1/2

 

Now let's commute the variance:

 

variance = [ ((0 - 1/2)² * 1/2) + ((1 - 1/2)² * 1/2) ]

 

I put extra parentheses to highlight the 2 terms being individually computed and then summed.

 

variance = [ (1/8) + (1/8) ] = 2/8 = 1/4

 

You may see the term "standard deviation" used. The "standard deviation" is

the square root of the variance.