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1 random v ec tor with mean µ y and v ar iance co v a1) where δ is the Dirac delta function This property of a Green's function can be exploited to solve differential equations of the form L u (x) = f (x) {\displaystyle \operatorname {L} \,u(x)=f(x)~} (2) If the kernel of L is nontrivial, then the Green's function is not unique However, in practice, some combination of symmetry , boundary conditions and/or other externally imposedNote that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero For example, y = x 2 4 y = x 2 4 is also a solution to the first differential equation in Table 41We will return to this idea a little bit later in this section

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Y ̔ R ̒ ʼnԂ̎ R ȕ\ G ߊ ؂ɂ i ` X ^ C ̉Ԃ Ă Ă ܂ BU(x;y) = F(x) G(y) 13 Conclusion Notice that where the solution of an ODE contains arbitrary constants, the solution to a PDE contains arbitrary functions In the same spirit, while an ODE of order mhas mlinearly independent solutions, a PDE has in nitely many (there are arbitrary functions in the solution!) These are consequences of theE 0 v v } r w r } x r r y { { E X C X G X ;

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1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !1 If X and Y are independent rv's then E(XjY)=E(X) Proof As we know, X and Y are independent if and only if fX;Y(x;y) = fX(x)fY(y) or, equivalently, fXjY(xjy)= fX(x) But then E(XjY =y)=åx xfXjY(xjy)=åx xfX(x)=E(X) 2 2 EE(g(X)jY)=E(g(X)) Proof Set Z = g(X) Statement (i) of Theorem 1 applies to any two rv's Hence, applyingX e B u E K b h g P X W W j N P X V X E J X e P X V U Z u X E A x j T E X X ^ E J t F X ^ E J t F P X X Q

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P Y T H A G O R E A N T R IA D S O F T H E F O R M X , X 1 , Z D E S C R IB E D B Y R E C U R R E N C E S E Q U E N C E S T W F O R G E T a n d T A L A R K IN L o ckh e e d M is s ile s & S p a ce C o m p a n y , S u n n y v a le , C a lifo rn ia T h e te rm P y th ag o rean T rip les o r T riad s is ap p lied to th o se in teg ersIf X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgf's Sometimes to stress the particular rv X, we write MŁy w A T V e B X z ł͓d b \ 񂵂Ă Ƃɂ A ҂ Ԃ̒Z k ͂ Ƃ A X ̂ q l ɑ΂ ăJ b g 鎞 Ԃ `30 ŁA l i ~( ō ) ƃ Y i u ɁI { p ̓v ̋Z Œ J Ɏ{ Ă ܂ B q l ɂ d オ ɂȂ 悤 A X ^ b t ꓯ A ͂ Ă ܂ ̂ŁA S B

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