00:01
Hello student, in the given question in part 1st we have to show that y upon theta is the pivotal quantity.
00:09
So, we need to demonstrate that the distribution of y upon theta does not depend on the parameter theta.
00:15
So, let's calculate the cumulative distribution function means cdf of y by theta.
00:24
So, the formula is f of y upon theta which is equals to probability of y upon theta is less than equals to t which is equals to probability of y is less than equals to t multiply by theta which is equals to integration of 0 to t 0 ,t multiply by theta into f of y of dy.
00:58
Now, substituting the probability distribution function of cdf of f of y is equals to 3 into theta minus y upon theta square.
01:13
So, we have the value as y is equals to f of y upon theta is equals to integration from 0 to t multiply by theta into 3 of theta minus y upon theta square into dy.
01:39
So, 3 which is equals to 3 upon theta square into integration of 0 multiply by 0 into t multiply by theta into theta minus y into dy which is equals to 3 upon theta square integration of theta y minus y square upon 2 which is equivalent evaluated from 0 to t of theta t multiply by theta which is equals to 3 upon theta square multiply by theta to the power 2 t minus theta square upon 2.
02:31
Simplifying this equation we get after simplification we get f of y upon theta which is equals to 3 upon theta square into bracket theta to the power 2 t minus t square multiply by theta square upon 2 which is equals to 3 t minus 3 by 2 into t cube.
03:02
Now, we observe that cdf of f of y upon theta does not depend on the theta therefore, y upon theta is pivotal quantity.
03:13
Therefore, we can observe that cdf of f of y upon theta does not depend on theta therefore, y upon theta is pivotal quantity...