# sol3 - Problem 1(a Keep in mind that we want to prove x2 x2...

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Problem 1 (a) Keep in mind that we want to prove n ( · ; 0 , 1) * n ( · ; 0 , 1) = n ( · ; 0 , 2) = 1 2 2 π e - x 2 2(2) = 1 2 π e - x 2 2(2) . (1) With this in mind: n ( · ; 0 , 1) * n ( · ; 0 , 1) = Z -∞ n ( x - y ; 0 , 1) n ( y ; 0 , 1) dy = Z -∞ 1 2 π e - ( x - y ) 2 2 1 2 π e - y 2 2 dy = 1 2 π Z -∞ e - ( x - y ) 2 + y 2 2 dy, at this point, we will multiply and divide the integrand by the exponential in (1), and this step will allow us to achieve the goal, which is precisely getting the expression in (1). Notice that this is permissible since the exponential in (1) only depends on x , which is a constant inside the integral: = 1 2 π e - x 2 2(2) Z -∞ e - ( x - y ) 2 + y 2 2 e x 2 2(2) dy = 1 2 π e - x 2 2(2) Z -∞ e - x 2 4 - xy + y 2 dy. (2) So far, only the usual rules of algebra have been used. By comparing the expressions (1) and (2), notice that it suffices to prove that the integral in (2) is equal to π . This is done as follows: Z -∞ e - x 2 4 - xy + y 2 dy = Z -∞ e - ( y - x 2 ) 2 dy = Z -∞ 1 2 e - z 2 2 dz, which we got by making the change of variables z = 2 ( y - x 2 ) , dz = 2 dy ; now by multiplying and dividing by π we get: = π Z -∞ 1 2 π e - z 2 2 dz = π, because the final integral is 1 by observing the integrand is the standard normal density. 1

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(b) The density of Z = X 1 + X 2 was computed in class and shown to be f Z ( z ) = λ 2 ze - λz 1 [0 , ) ( z ) . (Recall that it is a good habit to express the densities using this indicator notation, espe- cially if the domain is broken into two parts). On the other hand, f X 3 ( x ) = λe - λx 1 [0 , ) ( x ) . Therefore, by the convolution formula, the density of X 1 + X 2 + X 3 = Z + X 3 = X 3 + Z is given: Z -∞ f X 3 ( x - y ) f Z ( y ) dy = Z -∞ λe - λ ( x - y ) 1 [0 , ) ( x - y ) λ 2 ye - λy 1 [0 , ) ( y ) dy = λ 3 e - λx Z -∞ y 1 [0 , ) ( x - y )1 [0 , ) ( y ) dy = 1 [0 , ) ( x ) λ 3 e - λx Z x 0 ydy,
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