E[X(t—uy,) X(t—uz) X(t+T-¥) X(t+T—-V2)] 
= Ryx(u, — U2) Ryex(v1) —V¥2) + Ryx(t + uy —V1) Ryx(t + U2 -— V2) 
+ Ryx(tT + Uy —V2) Ryx(T + U2 -¥}). (11.39°) 
When this expression is inserted in Eq. 11.38, the term that comes from the first 
term of Eq. 11.39 is the same as m? in Eq. 11.33 and it cancels with the last term of Eq. 
11.38, — my. 
Then from the remainder 
Ryy({t) = J J h(s}) h(r1) Rxx(t + 81-11) 
+ ! J J J h2(uy, U2) h2(v1,¥2) Ryxx(t + uy—V1) Rxx(t + uz — v2) 
du,duzdv\dv2z 
h2(uy, u2)h2(V1, V2)Rxxx(T + Uy — V2)Rxx(T + U2 — V1) 
+ 
=< 
ie 
8 
se 
du,duzdv,dvz. (11.40) 
The terms h2(u;, U2), h2(vj, V2) are symmetrical for u;, v2 and v;, v2, and the scope of the 
integral is the same as for — © — + © for both the last two terms. Accordingly, the sec- 
ond and third terms are equal. Therefore, 
foo) 
Syy(@) == | Ryy(t) ec dt 
—-o 
1 
= Hy(w)| ? syxx(w) + | [x | expt-ilo-(o: +02) at 
—o 
X 2H2 * (@1,@2) H2(W1,02) SxA1)5xx (W2) dwidwp. (11.41) 
Here, using the relation 
x | ox [i{(@1+@2)-a@}t] dr =6{((@;+@2)-o} (11.42) 
6: Dirac’s delta function. 
305 
