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multiple choice
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Assuming that the random variables X and Y are independent and identically distributed, and both obey the uniform distribution on (0, 1), then among the following random variables that still obey the uniform distribution on the corresponding interval or area is ().
A. X 2 X^2X2
B. X + Y X+Y X+Y
C. X − Y X-Y X−Y
D. ( X , Y ) (X, Y) (X,Y )
【Correct answer: D】 -
It is known that the two-dimensional random variable (X, Y) obeys the uniform distribution on the area G, where G is surrounded by three straight lines xy=0, x+y=2, y=0, the incorrect one of the following options is () .
A. The density function of (X, Y) is f ( x , y ) = { 1 , 0 < y < 1 , y < x < 2 − y , 0 , others . f(x,y)= \begin{cases } 1,&0<y<1,&y<x<2-y, \\ 0,&others.& \end{cases}f(x,y)={ 1,0,0<y<1,others.y<x<2−y,
B. X的密度函数为 f X ( x ) = { x , 0 < x < 1 , 2 − x , 1 ≤ x < 2 , 0 , o t h e r s . f_X(x)= \begin{cases} x,&0<x<1, \\ 2-x,&1 \le x<2, \\ 0,&others. \end{cases} fX(x)=⎩⎪⎨⎪⎧x,2−x,0,0<x<1,1≤x<2,others.
C. 对0<y<1, P { X > 1 ∣ Y = y } = 1 3 P \{ X>1|Y=y \} = \frac {1}{3} P{ X>1∣Y=y}=31
D. 对0<y<1, P { X > 1 ∣ Y > y } = 1 2 P \{ X>1|Y>y \} = \frac {1}{2} P{ X>1∣Y>y}=21
【Correct answer: C】 -
Suppose two-dimensional random variable (X,Y)~N(0,1,1,1,0), then there is ().
A. P { X < Y } = 1 2 P \{ X<Y \} = \frac {1}{2}P{ X<Y}=21
B. P { X > Y } = 1 2 P \{ X>Y \} = \frac {1}{2} P{ X>Y}=21
C. P { X = Y } = 1 2 P \{ X=Y \} = \frac {1}{2} P{ X=Y}=21
D. P { X 2 + Y 2 < 1 } = 1 − e − 1 2 P \{ X^{2}+Y^{2}<1 \} =1-e^{- \frac {1}{2}} P{ X2+Y2<1}=1−e−21
【Correct answer: D】 -
Assuming that the random variables X and Y are both subject to normal distribution, the correct one of the following options is ().
A. X+Y must obey the normal distribution.
B. X+Y may not necessarily obey the normal distribution.
C. (X,Y) must obey the normal distribution.
D. XY must obey the normal distribution.
【Correct answer: B 】
Analysis:
When X and Y are independent of each other, X+Y obeys a normal distribution, and it is not necessarily true if they are not independent