Therefore, according to the probability characteristic function, f (t) = e [exp {jtx (t)}] = ∫ (-∞↗ +∞) exp {jtx} f (x) dx, ①.
Therefore, f (t) = e [exp {acos (wt+θ)}] = ∫ (-∞ ↗ +∞) exp {jtacos (wt+θ)} f (θ) dθ.
=[ 1/(2π)]∫(-π↗+π)exp{jtacos(wt+θ)}dθ=[ 1/(2π)]∫(-π+wt↗π+wt)exp { JTA cosy } dy。
According to the integral property, if f(t) is a periodic function with a period of t, then ∫ (-t/2+a ↗ t/2+a) f (t) dt = ∫ (-t/2 ↗ t/2) f (t) dt;
Therefore, f (v) = [1/(2π)] ∫ (-π ↗π) exp {jtacosy} dy = ∫ (-a ↗ a) exp {JTA} (dx)/(π √ a. -x？ ) , ②
Comparing ① and ②, the probability density function of X(t) is f(x)={ 1/(π√a? -x？ )，| x | & lta； 0, other.
According to the definition, x (θ) = e [acos (wt+θ)] = [1/(2π)] ∫ (-π ↗π) acos (wt+θ)} dθ = 0;
X (θ) = variance of a? e[cos(wt 1+θ)cos(wt2+θ)]= a？ [ 1/(2π)]∫(-π↗π][cos(wt 1+θ)cos(wt2+θ)]dθ
=(a？ /2) [cosw(t2-t 1)]。
2. Let t 1=t2=t, then X(θ) =(a? /2) ; So the covariance of the random variables X(t 1) and X(t2) =D[X(t)].
= e {[expected value of x (t)-x (θ)]? }=E [X？ (t)]-0=a？ /2.