Thanks. I feel good to know that the information was useful to you.

Glad you liked it! Do help the channel to reach out to others as well! Thanks once again!

Thanks for the appreciation. I am glad you find the content useful. Yes, more videos are in the making. Content is WIP. Happy viewing 😃

@@LearningPuree Thanks Sir. Looking forward to it :)

Yes Thanks. New video content is WIP.

Thanks. More are on the way.

The explanation will not change since the concept behind df does not change. You will have to look at the various variables in context of their equation. Explaining each variable will be a set of sessions on statistics.

Thanks

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It is good that you noticed! The summation was missing by design to keep the formula simple. It was also assumed that the previous tutorial on SD that has the complete formula was viewed. Thanks once again ...

Hello Niketan. Though the equation of covariance uses deviation of the 2 variables from their respective means, it needs to be understood that it is the "sum of product of the deviations" and not "sum of deviations multiplied with each other". Fundamentally, we are looking at how these variables "co-vary" and not "contra-vary". The primary assumption therefore being that they are dependent on each other (net effect). If they were independent then the cov(X,Y)=0. Intuitively, this would be clear from the fact that the sum of deviations for each variable would be 0. The explanation to your question requires some understanding of linear algebra and co-ordinate geometry/vectors. This is where the explanation becomes a bit technical. However, it should suffice to understand conceptually in simple terms that we are looking at an overall/net outcome of the "dependence". The net outcome being a single entity and therefore, same sample size.

@@LearningPuree sir I know that from Google search earlier we have 2n degrees of freedom. After estimate of mean of x and y we loose 2 df. Then after product of deviation we loose n-1. So net is n-1. But my query is why after product of deviation we loose n-1.

Niketan, this is exactly why I mentioned that the explanation requires a combined understanding of linear algebra, vectors and co-ordinate geometry. As mentioned earlier, we cannot look at x and y in isolation. It is the net effect of 2 vectors that is under study. This space may not be suitable for discussing the same.

@@LearningPuree sir thank you. With this i got intuition behind this. Consider a portfolio of x and y. Ie. (X+Y)=p. It is a combined portfolio of x and y independent. X and Y together form new variable. So we are concerned with new variable. Hence we loose only 1 degree of freedom. Mean of (x+y) ie one parameter estimated. We focus on covariablity . Means new variable of X+Y. Is it true?

Yes, this may be partially analogous to the example shared by you. However, I would like to point out that if x and y are independent they may not co-vary. They will vary independently of each other. In Statistics, when we consider independent events, we say that one does not affect the other. However, when covariance is established, the events are not independent. The outcome, covariance, is a measure of strength and direction of the combined interaction of the 2. Covariance is simply a measure of a 'probable pre-existing interaction between x and y'. It, in a way, may lay the foundation to examine if a cause-and-effect relationship exists between the x and y. So, in terms of portfolio, does some industry in a portfolio affect another industry in another portfolio and vice versa. Further, does that industry do this consistently, with what vigor and which direction? e.g. interaction or relationship of crude oil supply and price hitting an industry that use it as RM. Conversely, the industry/s may find a solution to crude oil and hit the price of crude. Since this measure is a singular outcome between the 2, it results in a single degree of freedom lost at a combined level. However, we are not 'creating' any new variable.

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