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I want to have some clarifications related to below question from Casella Berger

I want to have some clarifications related to below question from Casella Berger

Regarding $\phi_1(X_1),$ I understand that $\alpha$-value is 0.05 and hence we are looking for a value of $C$ for which $\phi_2(X_1,X_2)$ is also 0.05

When I plot $Y = X_1+X_2.$ I get uniform distribution of $(0,2),$ assuming $theta = 0.$ enter image description here

Hence, question reduces to value of $C$ so $Y(0,2) > C.$ Since this is a uniform shouldn't $C$ be equal to $0.95*2 = 1.9?$ However, correct answer is $C = 1.68$ (approx)

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    $\begingroup$ I have edited your question using JaX-notation. Please take a look to make sure I have not misunderstood your intended meaning. // Also because this is a textbook problem, please include the 'self-study' tag. $\endgroup$ – BruceET Jun 5 at 3:00
  • $\begingroup$ Thanks @BruceET for your help !! $\endgroup$ – Dataist Jun 5 at 3:10
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You have added cdfs, but the sum of cdfs (the thing you calculated) is NOT the cdf of the sum (the thing you need).

For example, try simulating the sum of two uniform random variates (say $10^4$ values or so) and see what a histogram (with say 100 bins) and an ecdf of that looks like (which will approximate the density and cdf in large samples)

Instead, the pdf of the sum is the convolution of the pdfs. If you then want a cdf, integrate that (though the required value can be obtained fairly easily by simple geometric reasoning/calculation).

[My suggestion is not to jump straight to writing code to try to solve a problem like this. Understand what you're doing first, then you won't waste time coding things that don't relate to what you're trying to achieve.]

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  • $\begingroup$ Thanks a lot @Glen_b. This helped. Correct pdf of Y (X1+X2) is a triangular distribution and not a uniform distribution. $\endgroup$ – Dataist Jun 5 at 3:45
  • $\begingroup$ Correct. Do you get the right answer for the upper tail critical value $C$? $\endgroup$ – Glen_b -Reinstate Monica Jun 5 at 4:21
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    $\begingroup$ Thanks @Glen_b, now I get the right answer of C = 1.684. I got it by solving $(2-C)^2 / 2 = 0.05$ $\endgroup$ – Dataist Jun 5 at 4:38

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