Question

(4) Let X,YX,Y be iid Uniform(−1,1) random variables. Find the density of Z=X+Y, and find the characteristic function of Z. By using the inversion formula deduce that

.∫0∞(sin⁡tt)2dt=π2.

The following ``answers'' have been proposed. Please read carefully and choose the most complete and accurate option.

(a) The characteristic function of X is sint/t. The characteristic function of Z is (sint/t)^2, which is integrable. If fZ(x) is the density of Z then

fZ(x)=12π∫−∞∞(sin⁡t/t)^2 e^−itx dt.

On the other hand, Z has the triangular density

fZ(x)=2−|x|/4,x∈(−2,2),

and zero otherwise. Hence, by taking x=0 we get

12=fZ(0)=12π∫−∞∞(sin⁡t/t)^2dt.

(b) The characteristic function of X is sint/t. The characteristic function of Z is (sint/t)^2, which is integrable. If fZ(x) is the density of Z then

.fZ(x)=12π∫−∞∞(sin⁡t/t)^2 e^−itx dt.

On the other hand, Z has uniform density over (−1,1). Hence, by taking x=0 we get

12=fZ(0)=12π∫−∞∞(sin⁡t/t)^2dt.

(c) The characteristic function of X is sinttsin⁡tt for t>0 and zero otherwise. The characteristic function of Z is (sint/t)^2 fort>0 and zero otherwise, which is integrable. If fZ(x) is the density of Z then

fZ(x)=12π∫−∞∞(sin⁡t/t)^2 e^−itx dt.

On the other hand, Z has the triangular density

f,fZ(x)=1−|x|/4,x∈(−1,1),

and zero otherwise. Hence, by taking x=0 we get

14=fZ(0)=12π∫0∞(sin⁡t/t)^2dt.

(d) The characteristic function of X is sinttsin⁡tt, for t>0. The characteristic function of Z is (sint/t)^2, for t>0, which is integrable. If fZ(x) is the density of Z then

fZ(x)=12π∫0∞(sin⁡t/t)^2 e^−itx dt.

On the other hand, Z has uniform density over(−2,2). Hence, by taking x=0 we get

14=fZ(0)=12π∫0∞(sin⁡t/t)^2dt.

(e) None of the above.

The correct answer is

Answer 1

The correct answer is (e).

For a, density of z is wrong

For b, z is not uniform on (-1,1)

For c, density of z is correct but f(0)=1/4 is wrong

For d, density of z is wrong.