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(3) Invert the test derived in (2) to find a (1-α)-Cl for λ. Problem 5. Suppose Yi, n 2 2. ,h are i.i.d. draws fron N(μ, σ*) with (1) Explain briefly why where s2 is the sample variance (unbiased for ơ2). Now suppose that the prior distribution on the pair (u, log a) is flat (Lebesgue in 2 dimensions), a priori, we work with respect (2) Find the posterior mean of μ and σ2, given the sufficient statistic (Y, s2). (3) Show that the posterior distribution of u is Y +st-i/Vn.


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Question: prove by induction 2^2 + 4^2 + 6^2 + ... + (2n)^2 = (2n)(2n+1)(2n+2)/6 ANSWER we will use induction on n base case : n=1 we have, 2^2 = 2*3*4/6 = 4 which is true inductive hypothesis let it be true for n = k i.e.,  2^2 + 4^2 + ... + (2k)^2 =   [(2k)(2k+1)(2k+2)]/6 inductive case let n = k+1 then we have 2^2 + 4^2 + .... + (2k)^2 + (2(k+1))^2 =   [(2k)(2k+1)(2k+2)]/6 + (2k+2)^2 =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = =(2k+2)*[(2k)(2k+1)/6 + (2k+2)] =(2k+2)*[ (4k^2+2k)/6 + (12k + 12)/6 ] =(2k+2)*[ (4k^2+14k+12)/6 ] = (2k+2)*[ (4k^2 + 8k + 6k + 12)/6 ] = (2k+2)*[ (4k(k + 2) +6(k+2))/6 ] = (2k+2)*[ (4k+6)(k+2)/6 ] =  (2k+2)*[ 2 (2k+3)(k+2)/6  ] =   (2k+2)*[  (2k+3)*2*(k+2)/6  ] =   (2k+2)*[  (2k+3)(2k+4)/6  ] = [(2*(k+1))(2*(k+1)+1)(2*(k+1)+2)]/6 replacing k+1 by m, we get replacing k+1 by m, we get [(2*m)(2*m+1)(2*m+2)]/6 this completes our proof b...

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