Requesting Intuitive Explanation to Covariance, Correlation and Standard Deviation First I would like to state that I am not from a mathematical background.
I am studying about change in price of products. So I have to understand about Correlation , Covariance and Standard Deviation intuitively. I have read lot of theoretical definitions today and little bit confused. 
I have taken the following data as example
When the price of Product X is Rs. 1,2,3,4,5 , the price of product Y is Rs. 3,6,9,12,15. 
I use Karl Pearson formula. 
The Covariance turns out to be 6 . I dont understand what this number 6 is. My understanding is covariance measures how much Y changes when there is a change in X. But here I thought Covariance should be 3. But it is 6 . So please explain what's wrong with my understanding. 
Here the Units of both X and Y is common only. I read that we divide the Covariance by Product SD of X and Y to arrive at a unit free value between X and Y. My first doubt is , what is the intuitive explanation of dividing the Covariance by Product of the SD of two variables. 
The correlation turns out to be 1 . My understanding of Correlation is it measures the strength and not the slope. When correlation is 1 it means , when X changes in one direction , Y always changes in the same direction . When Corrrelation is 0.9 it means 90 % of the time the variables move in the same direction.
And final doubt is regarding Standard Deviation. I understand SD as average distance of each variable from mean. So why don't we take the absolute value of differences from mean and average it. I heard this wont be accurate if the data's spread is high and I understand it intitively. But I dont get the derivation of formula mathematically.
I would be really really grateful if someone could explain the concept with numbers instead of theory ( would love if it is based on my example data) in a simple way to a non-math student like me. !!
 A: First, I will use some notation to make things easier:


*

*$SD(X) = $ standard deviation of $X$ 

*$SD(Y) = $ standard deviation of $Y$

*$Var(X) = $ variance of $X = SD(X) \times SD(X)$

*$Cov(X,Y) = $ covariance between $X$ and $Y$ 

*$Cor(X,Y) = $ correlation between $X$ and $Y$


Now that you have that down, I should explain what these things mean.  


*

*Standard Deviation is the measure of how far values deviate from their average value.  A low standard deviation indicates that data points tend to be close to the average.  A high standard deviation indicates that data points tend to be far away from the average.

*Covariance is the measure of "joint variability" between two variables (X and Y in this case).  Positive covariance means that when values of X increase, values of Y generally also increase.  Negative covariance means that when values of X increase, values of Y generally decrease.  Zero covariance means that when the values of X increase, this has no effect on Y.

*Correlation is a standardized way of thinking about covariance.


Now on to formulas, then on to your example.
$$
N = \text{number in sample from your population} \\
\mu_X = \text{average of X} \\
Var(X) = \frac{1}{N} \times \sum_{i=1}^N (X_i - \mu_X)^2 \\
$$
In your sample, for $N_X = 5$.  This means there are 5 observations of $X$ in your sample.  $\mu_X$ is the average of $X$, which is calculated like this:
$$
\mu_X = \frac{(1 + 2 + 3 + 4 + 5)}{N_X} = \frac{(1 + 2 + 3 + 4 + 5)}{5} = \frac{15}{5} = 3 = \mu_X
$$
Now, to calculate variance of X, we use the formula I stated above, which uses summation notation.  I will show you the calculation here:
$$ 
Var(X) = \frac{1}{5} \times 
\{ (1 - 3)^2 +  (2 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (5 - 3)^2\} \\
= \frac{1}{5} \times 
\{ 4 +  1 + 0 + 1 + 4 \} = \frac{1}{5} \times 10 = 2 = Var(X)
$$
Now that we have the variance of $X$, we can easily get the standard deviation of $X$, since we know that $[SD(X)]^2 = Var(X)$.  
$$
SD(X) = \sqrt{Var(X)} = \sqrt{2} = SD(X)
$$
Now you have the standard deviation of X.  We need to get $SD(Y)$ in the same way:
$$
\mu_Y = 
\frac{(3+6+9+12+15)}{N_Y} =  
\frac{(3+6+9+12+15)}{5} = 
\frac{45}{5} = 9 = \mu_Y
$$
$$
Var(Y) = \frac{1}{5} \times 
\{ (3 - 9)^2 +  (6 - 9)^2 + (9 - 9)^2 + (12 - 9)^2 + (15 - 9)^2\} \\
= \frac{1}{5} \times 
\{ 36 +  9 + 0 + 9 + 36 \} = \frac{1}{5} \times 90 = \frac{90}{5} = 18 = Var(Y)
$$
$$
SD(Y) = \sqrt{Var(Y)} = \sqrt{Var(Y)} = \sqrt{18}
$$
Now we move on to covariance, which uses the following formula:
$$
Cov(X,Y) = \frac{1}{N} \times \sum_{i=1}^N (X_i - \mu_X)\times(Y_i - \mu_i)
$$
This seems tricky, but here is the arithmetic:
$$
Cov(X,Y) = \frac{1}{5} 
\times \{ (1 - 3)(3 - 9) + (2 - 3)(6 - 9) + (3 - 3)(9 - 9) + (4 - 3)(12 - 9) + (5 - 3)(15 - 9)  \} \\
= \frac{1}{5} \times \{ (-2)\times(-6) + (-1)\times(-3) + (0)\times(0) + (1)\times(3) + (2)\times(6) \} \\
= \frac{1}{5} \times \{ 12 + 3 + 0 + 3 + 12 \} 
= \frac{1}{5} \times 30 = 6 = Cov(X,Y) 
$$
We are almost done!  Now we need to calculate correlation, which is just a standardized way to represent covariance that lets us compare two variables with different sized numbers (correlation of ear size to height; sure, they will be correlated, but height will vary a lot more than ear size because the units of height are much larger, causing a larger variance).
$$
Cor(X,Y) = \frac{Cov(X,Y)}{SD(X)\times SD(Y)}
$$
We know all of the above values, so we can calculate straight-away!
$$
Cor(X,Y) = \frac{6}{\sqrt{2} \times \sqrt{18}} 
= \frac{6}{\sqrt{2 \times 18}} = \frac{6}{\sqrt{36}}
= \frac{6}{6} = 1 = Cor(X,Y)
$$
This result makes sense, as we see that every time $X$ increases by 1, $Y$ increases by 3.  This happens for every movement, meaning that every time $X$ moves, $Y$ will respond linearly, in the same amount, in a perfectly correlated fashion.
There you have it, Variance 101.
