Let 𝑋 and 𝑌 be independent continuous random variables that are uniformly distributed on (0,1) . Let 𝐻=(𝑋+2)𝑌 . Find the probability 𝐏(ln𝐻≥𝑧) where 𝑧 is a given number that satisfies 𝑒^𝑧<2 . Your answer should be a function of 𝑧 .

Question

1)𝐏(ln𝐻≥𝑧)=
2)Let 𝑋 be a standard normal random variable, and let 𝐹𝑋(𝑥) be its CDF. Consider the random variable 𝑍=𝐹𝑋(𝑋) . Find the PDF 𝑓𝑍(𝑧) of 𝑍 . Note that 𝑓𝑍(𝑧) takes values in (0,1) .
𝑓𝑍(𝑧)=

I draw a grap of ln H, but e^z<2, meaning z<ln 2, which does not give any information in terms of ln𝐻≥𝑧.

Can anyone help solve this?
Thank you

Answer 1

Not sure if that's the right answer, but I solved by myself:
1) e^z/6
2) 1

Answer 2

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?

Answer 3

I got P(Y≥e^z/x+2) and don't know how to go further. It seems to be getting a function of z and x

Answer 4

From that point, P(Y≥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)))

Answer 5

a) 1-((e^z)*(ln(3/2)))
b) 1

Answer 6

fz = 1/sqrt(-2*sqrt(z*sqrt(2*pi)))

can anymone confirm ??

Answer 7

typo, should read: fz = 1/sqrt(-2*ln(z*sqrt(2*pi)))