Probability joint density

👨🏻‍🏫 學術台數學

[匿名]

2010-04-22 4:53 am

Please help, this is the question about the joint density.

Let D={(x,y) : 0<= x<=1 , y>=0 : 0<=x+y<=1}.Let two ransom variables X and Y be such that the random vector (X,Y) is uniformly distributed on D.

1>Find the joint density f(x,y).
2>Find P(X>Y+0.5)

更新1:

to nelsonywm2000 : sorry, i can't see the picture, can you post it again?

更新2:

i can see now. and could you teach me how to do fY(y) and fX/Y(x/y) [conditional density] as well ? thank you very much

更新3:

thank you veru much. i know i am troublesome but please help me for the last two. a) P(X>0.1) and P(Y>0.1) b) P(X>0.1|Y=0.2)

回答 (3)

用戶9112

2010-04-22 5:27 am

✔ 最佳答案

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圖片參考：http://img219.imageshack.us/img219/9158/44587046.png

2010-04-21 21:28:02 補充：
http://img219.imageshack.us/img219/9158/44587046.png

2010-04-21 22:59:31 補充：
For f_Y and f_(X|Y), please see:
http://img227.imageshack.us/img227/4774/60750202.png

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匿名

2010-04-24 7:09 pm

the link is only valid for 7 days

圖片參考：http://imgcld.yimg.com/8/n/HA00220463/o/701004210139513873386430.jpg

參考: http://www.yousendit.com/transfer.php?action=check_download&ufid=THE3Q1ZzNnk1aWJ2Wmc9PQ&key=8533676a4779eee1c77f49873fe4580249df5cfb&bid=OHo2QmthU1A4NVZFQlE9PQ

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myisland8132

2010-04-22 5:41 am

1 Since the random vector (X,Y) is uniformly distributed on D
f(x,y)=k
Integrate it ∫ ∫ k dydx=1
∫ k (1-x) dx =1
k(x- x^2/2) | [0,1] =1
k/2=1
k=2

So f(x,y)=2 where D={(x,y) : 0<= x<=1 , y>=0 : 0<=x+y<=1}

2 For X > Y+0.5
Since y+0.5=1-y=>y=0.25
So the region becomes 0<= y <= 0.25, y + 0.5 <= x <= 1 - y
∫ ∫ 2 dxdy
= ∫ 2[(1-y)-(y+0.5)]dy
=2 ∫ (0.5-2y) dy
=y-2y^2 | [0,0.25]
=0.25-2(0.25)^2
=0.125

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收錄日期: 2021-04-23 23:23:17
原文連結 [永久失效]:
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