Question
Let π and π be independent continuous random variables that are uniformly distributed on (0,1) . Let π»=(π+2)π . Find the probability π(lnπ»β₯π§) where π§ is a given number that satisfies π^π§<2 . Your answer should be a function of π§ .
1)π(lnπ»β₯π§)=
2)Let π be a standard normal random variable, and let πΉπ(π₯) be its CDF. Consider the random variable π=πΉπ(π) . Find the PDF ππ(π§) of π . Note that ππ(π§) takes values in (0,1) .
ππ(π§)=
I draw a grap of ln H, but e^z<2, meaning z<ln 2, which does not give any information in terms of lnπ»β₯π§.
Can anyone help solve this?
Thank you
1)π(lnπ»β₯π§)=
2)Let π be a standard normal random variable, and let πΉπ(π₯) be its CDF. Consider the random variable π=πΉπ(π) . Find the PDF ππ(π§) of π . Note that ππ(π§) takes values in (0,1) .
ππ(π§)=
I draw a grap of ln H, but e^z<2, meaning z<ln 2, which does not give any information in terms of lnπ»β₯π§.
Can anyone help solve this?
Thank you
Answers
yyyyz
Not sure if that's the right answer, but I solved by myself:
1) e^z/6
2) 1
1) e^z/6
2) 1
prego
a)Any hints...did you integrate from the expression to 1 for both x and y?
your pdf does not integrate to 1
b)1?
your pdf does not integrate to 1
b)1?
Anonymous
I got P(Yβ₯e^z/x+2) and don't know how to go further. It seems to be getting a function of z and x
Anonymous
From that point, P(Yβ₯e^z/x+2), you can double integrate, and you result in the integration from 0 to 1 of 1-((e^z)-(x+2)). Therefore the result is:
1-((e^z)*(ln(3)-ln(2)))
resulting in:
1-((e^z)*(ln(3/2)))
1-((e^z)*(ln(3)-ln(2)))
resulting in:
1-((e^z)*(ln(3/2)))
yare
a) 1-((e^z)*(ln(3/2)))
b) 1
b) 1
fz = 1/sqrt(-2*sqrt(z*sqrt(2*pi)))
can anymone confirm ??
can anymone confirm ??
typo, should read: fz = 1/sqrt(-2*ln(z*sqrt(2*pi)))