???即： FY(y)?????0,y?0y,0?y?1 21,y?1?e?x,x?032.解： x~e(1)~fX(x)??
?0,x?0 Y＝ax+β
FY(y)?p(Y?y)?p(?X???y)?p(X?'fY(y)?Fy(y)?FX('y???
)?Fx(y???)
y???)?fX(y???)?1?
??1???y???e,＝???0,?(y??)y???0??1e?,＝????其他0,?1y?? 其他FY(y)?Fx(y???)?1?ey???y???
1?(y??)???,??1?e?0,?1?(y??)??0＝?1?e?,???0,其他?y??
其他
?1?33.解： 令X:直径 f(x)~?b?a??0 Y:体积 Y?a?x?b其他
4x?（)3 32baEY??g(x)f(x)dx??????ba4x31411?()?dx????x432b?a3b?a32?11??(b4?a4)24(b?a)
?1(b?a)(b2?a2) 2434.解：
FY(y)?p(Y?y)?p(?y?x2?X?y)?p(X??2?y)??20y1??e?dx
?x ???20e?d(?)??ex??x?y?20
yy???2?2???e?1??1?e . (y?0）
??所以：
21
??y F?2Y(y)??1?e,y?0
??0,y?0所以： dFyY(y)1dy?fy)?2e?Y(2 (y?0）
所以：
?yg(y)??1??fY(y)?22e,y?0 所以：Y~?2(2)
??0,y?0
.35.解 X~e(2)
所以： f?2e?2x,x?0?1?e?2x,x?0X(x)?? F?0,x?0X(x)??
?0,x?0FXY(y)?p(Y?y)?p(1?e?2?y)?p(e?2X?1?y)?p(?2X?ln(1?y)) ?p??X??1?1?2ln(1?y)???FX(?2ln(1?y))
??2(?1?2ln(1?y)??1?e,0?y?1??1,y?1 ??1?(1?y),0?y?1?y,0?y?11,y?1 ???1,y?1 ???0,y?0?0,y?0?0,y?0????y,0?y?1F(y)???1,y?1 f?1,0?y?1Y?Y(y)???0,y?0?0,其他
?1?12?(lnx??)36.解：由已知参考(p)知f????(lnx),x?0?2?266~67X(x)e,?x ???0,x?0?2??x?0,当前价格X0?10元。 (lnx?u)2 EX????2?20x?fX(x)dx????102??e?dx
依据 p67-eg2.31 EX?e???22?15?e2???2?152
??e?2?1?4?e?2229225?225
22
x?0x?0
DX?e2???(e??1)?4
229?2??ln??225??225???ln?229?2(lnx?u)?1??ln229?ln225?2?2?,x?0 其中 fX(x)??2??xe1??ln225?ln229?0,x?02?222
连续复合年收益率 r=lnx-lnx0=lnx-ln10
FR(r)?P(R?y)?p(lnX?ln10?y)?p(lnX?y?ln10)?p(X?10?ey)
??10?ey12??x0e?(lnx??)22?2dx lnx??)2?lnx??2?????所以：
10?ey10?2?e?(lnx??)22?2d(2) 令?v
ln10?y????22?e?vdv?2??0(ln10?y??2?)
FR(y)?2??0(y?ln10??2?)
fr(y)?Fr'(y)????
注：对数正态分布与对数正态分布的矩，包括中心矩，原点矩等，如EX，DX均不作要求属于超纲内容，Black－Scholes
期权定价公式一般是作为研究经济现象工具也属于超纲内容，因而本题不作要求。 37.证明：
FY(y)?P(Y?y)?p(X?y)?p(X?y2) 显然当y≤0时,P(X?y)?0，所以 fY(y)?0 y＞0时，FY(y)?p(X?y)?y22?y20f(x)dx
dFY(y)d?0f(x)dxfY(y)???2y?f(y2)（复合函数求导方法）
dydy所以：
?2y?f(y2), fY(y)??0,?y?0 y?038.解:X密度函数fX(x), Y=ax+b
23
y?b FY(y)?p(Y?y)?p(aX?b?y)?p(X?)??afX(x)dx , a?0
??adFY(y)d?fY(y)??dy固当a＞0时， fY(y)?y?ba??y?bfX(x)dxdy?1y?b?fX() aa
1y?b?fX() aa1y?b) 当a＜0时，fY(y)???fX(aa
习题三答案
1. 证明：F?x2,y2??F?x1,y2??F?x2,y1??F?x1,y1? ?P?x1＜x?x2,y?y2??P?x1＜x?x2,y?y1?
?P?x1＜x?x2,y1＜y?y2??0由概率的非负性，知上式大于等于零，故得正. 2. 解：①， x2 x1 0 1 2 ０ 1/56 10/56 10/56 １ 5/56 20/56 10/56 PX2?x2j 3/28 15/28 5/14 ??P?X1?x1i? 3/8 5/8 1 ②. P?X1?0,X2?0???1?P{x2?0|x1?0}?P?x1?0?
?1??1??C27??3203205????821?8?56?14 ?P?X1?X2??P?x1?0,x2?0??P?x1?1,x2?1??21 56P?x1x2?0??1?P?x1x2?0??1?P?x1?1,x2?1??P?x1?1,x2?2?
?1?5102613??? 145656283. 解：①由概率的性质
??又
??0??0?x0?????3x?4yf?x,y?dxdy??0kedxdy?1?k1?12 01????x?3x?4yg?x,y?dxdy??0?0k2edxdy?1?k2?21
????0② fX?x?? fY?y???????3x?4y01kedy?k1e?3x????3x?4y01kedx?k1e?4y1?3e?3x x＞0 41??4e?4y y＞0 324
当y＞0时 gX?x??当x＞0时 gY?y????x0k2e?3x?4ydx?21?3xe?e?7x x＞0 4????yk2e?3x?4ydx?7e?7y y＞0
4. 解：①
?1?,0?x?2，0?y?1f?x,y???2
?0,其他?
?xy?,0?x?2，0?y?1xyF?x,y???01/2dxdy??2 ?0?0,其他?
②
fX?x???101121dy? 0?x?2；fY?y???0d?1，0?y?1
222x0?x?2?1?y?xyFX?x???0dx??2 ；FY?y???01dx??
2其他0??0? ③
0?y?1
其它
Py＜x?2???12111?y＜x22dxdy?1??0?x22dxdy?3
5.
?2f?x,y???
?00?x?y?1
其它0?x?1其它其它
fX?x???x1?2??1?x?2dy??
?0?2y2dx??
?0
fY?y???y0?y?1
00,其它??x?y?1?,0?x?2,1?y?2?0?2其它?1???fx,y?x? 6.①F?x,y??? ,0?x?2,y＞2??2?20?x?2,1?y?2?y?1,x＞2,1?y?2?1,x?2,y＞2?②
G?x,y??PX2?x,Y2?y?P?x＜X?x,?y＜Y?y?(xy?x)/2,0?x?4,1?y?4?0，x?0,或y?1??x G(x,y)??，0?x?4,y?42?y?1,x＞4,1?y?4??1,x＞4,y?4?????
讨论如下：
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