Suppose yt xt et where et are normal white noise with mean 0...

Suppose Yt = Xt + et, where {et} are normal white noise with mean 0 and variance σ^2. The {Xt} process is a stationary AR(1) defined by Xt = φXt−1 + Zt, where {Zt} is a zero mean normal white noise process with variance σ^2. Assume that Zt is independent of Xt−1, Xt−2, ... Assume additionally that E(etZs) = 0 for all t and s. (a) Show that {Yt} is stationary and find its auto-covariance function, γk. (b) Show that the process {Ut}, where Ut = Yt − φYt−1 = (1 − φB)Yt, has nonzero correlation only at lag 1.

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00:01 This question we have a big that containing 30 balls so by containing 30 balls that number is 1 to 30 so here we can write n of s which is equal to 30 here and the sample size s which is equal to we can write 1 2 3 to 30 1 2 3 to 30 and that is denoted the event is denoted event so we can write 24 6 8 to 10 we can write n of a which is equal to 15 9 of which in 5 that is equal to 15 and p which is equal to 5 for the probability for the success so we can write p of a which is equal we can write n owe divided by we can write n of s which is equal we can write 50 divided by 30, which is equal to 1 divided by 2.

00:50 Now here for the failure here for the failure that p which is equal to note e1 which is equal we can write here here 1 minus 1 divided by 2 which is equal to answer is 1 divided by 2 now we have 2 balls we have 2 balls that drawn from the bag which is a replacement so here we can write a the a the first one here that b which is equal we can add p of two successes we have two successes so here we can write p of first ball a 1 first wall e1 multiple by we can write multiply by and of second wall even second wall even and you're simply fine we can write one one divided by two multiply one divided by two which is equal answer is 0 .25 0 .25 now here second one is that for a p of exactly one success exactly one success which is equal you can write p of first a one wall first wall e1 and we just multiply by p multiplied by second wall second wall not even here you can write second one not a one plus we can write first board not even not even and this value is multiplied by v of second wall e1 here you can write second wall event so here we are substituted value so you can write 1 divided by 2 1 divided by 2 plus we can write 1 divided by 2 here we can write 1 divided by 2 we just simply find we can write here answer is 0 .5 so this is the value is 0 .5 now at p at least 1 success in the now in the third one in the third we know that p at least 1 success at least 1 success which is equal to we can write 1 1 minus v of no success and here you can do one minus be of first world is not even so we can write 1 divided by 2 multiple liby we can write v0 1 that is 1 divided by 2 we just simplifying we can write here answer is 1 divided by 1 minus 4 now we just again simplify and we can write 3 divided by 4 so this is the answer and we just simply find we can write here answer is 0 .75 so this is the answer so this is the answer let's find for the fourth one.

03:56 So the fourth one is that p of no success, p of no success, which is equal we can write p of not even, here first one is not even, that multiple is not even, that multiple above we can write, our second one is not even, we just simply find we can write here answers, 1 divided 4 which is equal to answer is 0 .25...

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