Introduction to Cramer-Rao Lower Bound

In this lesson, we’ll overview the concept of the Cramer Rao lower bound for the MSE for the estimation of a deterministic, but unknown parameter, and we show how to compute and interpret this bound for a few simple examples.

[@zemichaelseltan1719]

You made my day, thanks a lot.

[@alimuqaibel7619]

Very well presented with excellent examples. Thank you.

[@hazelndlovu3996]

could you do a video on biased and unbiased estimators please

[@dragosmanailoiu9544]

Should get more views

[@iaaan1245]

amazing video!

[@sohummisra8969]

In the second example shouldn't the solution for sigma_hat = (pi+2)sigma^2/2 instead of (pi-2)sigma^2/2?

[@timothyschulz9853]

No, it should be (pi-2)sigma^2/2.

[@sohummisra8969]

@@timothyschulz9853 can you explain how? I have considered zero mean as mentioned in the example. So E[X]=0 and according to the solution, E[X^2]= σ^2. So, using these, I get a positive sign instead of a negative.

[@sohummisra8969]

For an unbiased estimator E[σ']=σ, where σ' is sigma hat. So, E[(σ'-σ)^2] = E[σ'^2 + σ^2 -2σ*σ'] = (π/2)E[X^2] + E[σ^2] - 2σE[σ'] = (π/2)σ^2 + σ^2 - 2σ^2 = your solution.
However, if I consider E[σ'] = sqrt(π/2)*E[X], this term goes to 0 as E[X]=0. Can you please explain as to why the second approach is incorrect?

[@MegaSkuper]

in you binomial example, shouldnt CRLB be multiplied by N (number of observations)?

[@alexakelly8233]

In the second example, how does In(1/(sqrt(2pi sigma)) ) = 1/sigma^2 ? Thanks shouldn't it be -In(1/(sqrt(2pi sigma)) ?

[@dominicj7977]

there is a sigma^2 on top of e

[@dominicj7977]

2nd example it should be sigma hat = (pi+2)sigma^2/2

[@yannlegrands6939]

It would be better if there is not background music at the beginning, which is too loud.

[@victorkipngetich5653]

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