Question

Find P(Y >1|Y<=2). Y is a continuous random variable. F(y) = {0 when y>0} and {1-e^(-y^2) when y>=0}. f(y) = {0 when y>0} and {2y*e^(-y^2) when y>=0}

I know that P(Y>1|Y<=2) = [P(Y<1 AND Y <=2)]/P(Y<=2)]

I am confused at how to integrate the >1 part. is that from 1 to 2 or since it is not >= can we not just integrate from 1 to 2?

Answer 1

on the second paragraph that should be a P(Y>1 not P(Y<1

Answer 2

also, I was wondering is P(Y>=.5 AND Y>=.3) the same as P(Y>=.5)

Answer 3

If you guys have some time I would appreciate any help. thanks in advance.
@KingGeorge
@satellite73

Answer 4

i am confused by the inequalities, but may be it is because it is late

Answer 5

but also i am confused why you have to integrate, you have \(F\)
maybe i am just confused and should shut up, but can't you compute
\[F(2)-F(1)\]?

Answer 6

do you know if P(Y>1 AND Y <=2) = P(1<Y<=2)?

Answer 7

yes

Answer 8

but i don't understand the \(y>0\) and \(y\geq 0\) part

Answer 9

ok, then since its continuous, I think we can say
P(1<Y<=2) = P(1<Y<2)

Answer 10

yes

Answer 11

the y>0 and y>=0 is different conditions for the F(y) function. I think it is the "parameters" for the range of the function.

Answer 12

it is
\[F(y) = \left\{\begin{array}{rcc}
0 & \text{if} & y<0 \\
1-e^{-y^2}& \text{if} & y\geq 0
\end{array}
\right.\]

Answer 13

ok then let me shut up because i am confused by that

Answer 14

yes. I'm pretty bad at writing it in nice formatting.

Answer 15

i don't understand how it can be one condition for \(y>0\) and another for \(y\geq 0\)

Answer 16

that didn't make sense to me either. I think my prof might have made a mistype for the 0<y and it should have been 0>y.

Answer 17

i certainly hope so because it doesn't make sense in any context to me, probability or otherwise

Answer 18

F(y) is the probability distribution function, while f(y) is the probability density function. for the continuous random variable Y

Answer 19

i would not bet money on my answer, but my guess is it is
\[\frac{F(2)-F(1)}{F(2)-F(0)}\]

Answer 20

for the numerator of the conditional probability, I am integrating form 1 to 2 over the function \[2y*e ^{-y^2}\]

Answer 21

*from

Answer 22

i thought you were given \(F\)

Answer 23

no matter you will get the same thing

Answer 24

since the anti derivative is \(-e^{-y^2}\)

Answer 25

i am getting .35/.35 which equals 1. Now that i think about it it makes a lot of sense. we are taking the probability of everything above 1 and the probability of everything less than or equal to 2 so that would be the probability over all possible values which would be 1.

Answer 26

less = *else

Answer 27

i get
\[\frac{e^{-1}-e^{-4}}{1-e^{-4}}\]

Answer 28

probably more elegantly written as
\[\frac{e^3-1}{e^4-1}\]

Answer 29

yeah, that's what i get too. let me check my algebra again.

Answer 30

yeah i messed up on my algebra for the denominator. instead of .35 it should have been .98.

Answer 31

or if you want to be real fancy write
\[\frac{1+e+e^2}{1+e+e^2+e^3}\]

Answer 32

since it's a probability, I have to write it as a decimal. I get .357 when i divide.

Answer 33

such a nice answer
such an ugly decimal

Answer 34

that's true.

thank's for all the help. I really appreciate it. now off to sleep...