For the geometric distribution p(y) y (1 ), y 0, 1, 2, , show that the tail method for constructing a confidence interval i e , equating P(Y y) and P(Y y) to 2 yields ( 2) 1 y , (1 a 2) 1 (y 1) Show that all between 0 and 1 a 2 never fall above a confidence interval, and hence the actual coverage p...

The cdf is Fy 1 y1 Equating it to 2 and solving for yields the upper endpo ...

For the geometric distribution p(y) = y (1 ), y = 0, 1, 2,... , show

Question:

For the geometric distribution p(y) = π^y(1– π), y = 0, 1, 2,... , show that the tail method for constructing a confidence interval [i.e., equating P(Y ≥ y) and P(Y ≤ y) to α/2] yields [(α/2)^1/y, (1 – a/2)^1/(y+1)]. Show that all π between 0 and 1 – a/2 never fall above a confidence interval, and hence the actual coverage probability exceeds 1 – a/2 over this region.

Distribution
The word "distribution" has several meanings in the financial world, most of them pertaining to the payment of assets from a fund, account, or individual security to an investor or beneficiary. Retirement account distributions are among the most...

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