Consider the following MA(2) process yt = ut + ?1ut-1 +...

Consider the following MA(2) process y_t = u_t + ?_1u_{t-1} + ?_2u_{t-2} where u_t is a zero mean white noise process with variance ?^2. a) Calculate the mean and variance of y_t. b) Derive the autocorrelation function for this process. c) If ?_1 = -0.5 and ?_2 = 0.25, sketch the ACF of y_t

Transcript

00:01 In the given question we are having our equation as y t is equal to u z plus theta 1 u t minus 1 added by theta 2 t u this is sorry u with the base t minus 2 so here it is given u t is a non 0 0 mean of white noise process with the way which is sigma squared.

00:38 So now in the first part of the question first of all we will take our expected variance as y t or the equation as e u t plus theta 1 u t minus 1 added by theta 2 u t minus 2 that is equal to e of u t added by theta 1 a of u t minus 1 for the second term then for the last term here theta 2 is the constant value and now we will have here u t minus 2 so by using this formula where a is the constant so now we will get here b by t is equal to 0 plus theta 1 multiplied by 0 plus theta 2 multiplied by 0 which is equal to 0 here u t is non -zero mean process so now we will find our variance or pi t is equal to variance of u t plus theta 1 u t minus 1 plus theta 2 u t minus 2 and again with the same way variance of first u t plus theta 1 square variance of u p minus 1 plus theta 2 square variance of u t minus 2 because here u t is again unknown sorry this is a white now white noise process so then we will use here variance of a x is equal to a square variance of x where a is again constant so from here we will get variance of white t is equal to sigma square plus theta 1 square sigma square plus theta 2 square sigma square so we can take sigma common we will get here 1 plus theta 1 square plus theta 2 square so here is the solution for the first part of the question for variance of 5t we are having our equation this is answer for a part now let's move on to the second part of the question so in the part b we will take the auto collagration function auto corracation function so first we need to find here auto covariance function so this is gamma k is equal to covariance of y t comma y t plus k and here the k is time lag so therefore first of all we will find gamma not which is covariant of y t comma y t is equal to variance of y t and its value we have pointed in the first part as theta 1 squared plus teta 2 square multiplied by sigma square so now the value for y gamma 1 with the same way from the formula of covariance of y t y t plus 1 is equal to covariance of u t plus theta u t minus 1 plus theta 2 u t minus 2 then comma next value is utt plus 1 plus theta 1 u t plus theta 1 plus theta 2 utt minus 1 so here is the complete values then we know that the white noise process is a sequence of uncolidated random thus we can say here covariance of ui comma uj which is equal to 0 and here value of i does not equal to j this is the condition when i does not equal to j and then when we use our covariance formula as x or w plus x comma y plus z is written as covariance of w comma y covariance of w comma z added by covariance of x comma y added by covariance of x comma z so from here we will get gamma 1 as theta 1 covariance of u t comma u t again added by theta 1 theta 1 theta 1 o variance of u t minus 1, ut minus 1.

06:47 So by using our formula where c is constant, again, the value will be for comma 1 is theta 1 sigma squared plus theta 1, theta 2 sigma squared.

07:02 So this can be written as after taking sigma common, theta 1 plus theta 1, theta 2, sigma square and in the same way we can find our value for gamma 2 which is theta to sigma square and now gamma k will be equal to 0 and here k is greater than or equal to 3 because no more common variables in the expression of y t and y t plus k can be finded here so finally our value for ro k gamma k divided by gamma naught and this way row of p is equal to gamma not divided by gamma not which will be equal to 1 so in the same way p 1 will be as gamma 1 by gamma not in written as theta plus theta 2 divided by 1 plus theta 1 plus theta 2 2 after simplification sigma and sigma will be cancelled out we will have this really similarly root 2 will be as gamma 2 divided by gamma and our final value will be as theta 2 divided by 1 plus theta 1 square plus theta 2 square and our value for row k which is equal to 0 for k is greater than equal to 3 so here is the solution for b part we have find it the condition now let's move on to the next part which is part c in the c part if theta is equal to theta 1 is equal to minus 0 .5 and here theta 2 value is 0 .25 then row 1 is equal to 1 .2 will be equal to as when we calculate we will get here minus 10 by 21 which is equal to minus 0 .4 -7619 similarly this is for row 1, row 2 is equal to 4 by 21, 0 .190 .48 is the answer.

09:29 And here, pk is equal to 0 .4...

Question

Consider the following MA(2) process y_t = u_t + ?_1u_{t-1} + ?_2u_{t-2} where u_t is a zero mean white noise process with variance ?^2. a) Calculate the mean and variance of y_t. b) Derive the autocorrelation function for this process. c) If ?_1 = -0.5 and ?_2 = 0.25, sketch the ACF of y_t

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