Any of you guys good at probability? I may as well shoot for it, so here's the problem:
In medical imaging such as computer tomography, the relationship between detector readings Y and body absorptivity X is given by Y=eX. Determine the density function of Y when X has a normal distribution. (It's the standard normal, where the mean is 0 and the standard deviation is 1)
I keep getting this nasty mess that looks nothing like the solution.
For reference, the solution is fY(y)=(2/Ï€)1/2ye-(y^4)/2, y>0
Edited, May 4th 2011 7:43pm by Sweetums
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I hear alma is good.
If you were looking for dY/dx I could tell you that it's e^x. I took 3 courses of calculus and a course of differential equations but I never took anything in probability so I have no idea what a probability density function is.
Actually, what I am finding online, is that the normal distribution is
f(x) = 1/sqrt(2*pi)e^-(x^2)/2
maybe if you used the formula I found here of:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
That's all I can really think of.
Actually, what I am finding online, is that the normal distribution is
f(x) = 1/sqrt(2*pi)e^-(x^2)/2
maybe if you used the formula I found here of:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
That's all I can really think of.
Hyolith wrote:
If you were looking for dY/dx I could tell you that it's e^x. I took 3 courses of calculus and a course of differential equations but I never took anything in probability so I have no idea what a probability density function is.
Actually, what I am finding online, is that the normal distribution is
f(x) = 1/sqrt(2*pi)e^-(x^2)/2
maybe if you used the formula I found here of:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
That's all I can really think of.
Actually, what I am finding online, is that the normal distribution is
f(x) = 1/sqrt(2*pi)e^-(x^2)/2
maybe if you used the formula I found here of:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
That's all I can really think of.
the Y=e^X is pretty much Pr(Y≤y) = Pr(e^X ≤ y) = Pr(X ≤ log(y)) which is Fx(log(y)) (capital and lowercase f are two different functions, with the density being the derivative of the distribution function)
The problem is that f_x(log(y)) is a really nasty function with a log^2(y) and you can't really cancel that out with e and I'm stuck. The solution has no logs in it, and the distribution function of a normal distribution is also a nasty function I don't particularly understand.
Edited, May 4th 2011 10:40pm by Sweetums
What's the probability I read beyond the first line in the first post?
"I have lost my way
But I hear a tale
About a heaven in Alberta
Where they've got all hell for a basement"
But I hear a tale
About a heaven in Alberta
Where they've got all hell for a basement"
Iamadam the Malefic wrote:
What's the probability I read beyond the first line in the first post?
Sweetums wrote:
Not looking for dY/dX. I was mostly hoping that someone would know this because the density function for a normal distribution is pretty nasty (maybe you should have TeX support, Zam!) E[X]=0 (that's mu) and the standard deviation is 1, it would only be sqrt(1/2Ï€) * e^(-(x^2)/2))
the Y=e^X is pretty much Pr(Y≤y) = Pr(e^X ≤ y) = Pr(X ≤ log(y)) which is Fx(log(y)) (capital and lowercase f are two different functions, with the density being the derivative of the distribution function)
The problem is that f_x(log(y)) is a really nasty function with a log^2(y) and you can't really cancel that out with e and I'm stuck. The solution has no logs in it, and the distribution function of a normal distribution is also a nasty function I don't particularly understand.
Edited, May 4th 2011 10:40pm by Sweetums
the Y=e^X is pretty much Pr(Y≤y) = Pr(e^X ≤ y) = Pr(X ≤ log(y)) which is Fx(log(y)) (capital and lowercase f are two different functions, with the density being the derivative of the distribution function)
The problem is that f_x(log(y)) is a really nasty function with a log^2(y) and you can't really cancel that out with e and I'm stuck. The solution has no logs in it, and the distribution function of a normal distribution is also a nasty function I don't particularly understand.
Edited, May 4th 2011 10:40pm by Sweetums
I'm assuming with you say log(y) you mean the natural log (ln)? I'm assuming you're trying to get X by itself which means you would have to take the ln() of both side, from what I remember.
No offense Hyolith, but you're talking some pretty basic stuff here that I'm 99% sure Sweetums has already covered.
Can you post the furthest equation you've got? Maybe there's algebraic manipulations to get you to the answer.
Can you post the furthest equation you've got? Maybe there's algebraic manipulations to get you to the answer.
Wiki tells me normal distribution is (1/sqrt(2pi)) * e^(-.5x^2). It's either you need to integrate that, or you need to replace x with that in y=e^x and integrate that, or you need to set x = (1/sqrt(2pi)) * e^(-.5z^2) and double integrate y = e^x with respect to z first, then x.
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
Dear people I don't like: 凸(â—´―｀â—)凸
It's 8 + 4.
Hyolith wrote:
Sweetums wrote:
Not looking for dY/dX. I was mostly hoping that someone would know this because the density function for a normal distribution is pretty nasty (maybe you should have TeX support, Zam!) E[X]=0 (that's mu) and the standard deviation is 1, it would only be sqrt(1/2Ï€) * e^(-(x^2)/2))
the Y=e^X is pretty much Pr(Y≤y) = Pr(e^X ≤ y) = Pr(X ≤ log(y)) which is Fx(log(y)) (capital and lowercase f are two different functions, with the density being the derivative of the distribution function)
The problem is that f_x(log(y)) is a really nasty function with a log^2(y) and you can't really cancel that out with e and I'm stuck. The solution has no logs in it, and the distribution function of a normal distribution is also a nasty function I don't particularly understand.
Edited, May 4th 2011 10:40pm by Sweetums
the Y=e^X is pretty much Pr(Y≤y) = Pr(e^X ≤ y) = Pr(X ≤ log(y)) which is Fx(log(y)) (capital and lowercase f are two different functions, with the density being the derivative of the distribution function)
The problem is that f_x(log(y)) is a really nasty function with a log^2(y) and you can't really cancel that out with e and I'm stuck. The solution has no logs in it, and the distribution function of a normal distribution is also a nasty function I don't particularly understand.
Edited, May 4th 2011 10:40pm by Sweetums
I'm assuming with you say log(y) you mean the natural log (ln)? I'm assuming you're trying to get X by itself which means you would have to take the ln() of both side, from what I remember.
Majivo wrote:
No offense Hyolith, but you're talking some pretty basic stuff here that I'm 99% sure Sweetums has already covered.
Can you post the furthest equation you've got? Maybe there's algebraic manipulations to get you to the answer.
Can you post the furthest equation you've got? Maybe there's algebraic manipulations to get you to the answer.
Possibly, but the annotations for log and natural log are two completely different things.
Looking at it from a numbers standpoint:
log(20) = 1.30
while
ln(20) = 2.99
Looking at a derivative standpoint:
dy/dx of say log(x) assuming base 10 is 1/(x ln(10))
while
dy/dx of ln(x) is just 1/x
So it does have an impact on the answer, depending on the base of course because the ln(e) = 1
EDIT: Sweetums, answered it.
Edited, May 5th 2011 11:55am by Hyolith
Deadgye wrote:
Wiki tells me normal distribution is (1/sqrt(2pi)) * e^(-.5x^2). It's either you need to integrate that, or you need to replace x with that in y=e^x and integrate that, or you need to set x = (1/sqrt(2pi)) * e^(-.5z^2) and double integrate y = e^x with respect to z first, then x.
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
The density function fX(x) is given but you need to find fY(y), so you pretty much take the inverse of Y=eX so you can put the distribution function in terms of X.
FY(y) = Pr(Y≤y) which is equivalent to Pr(eX≤y) and then you take the log so you get FY(y) = FX(log(y))
d/dx (FX(log(y))) = fX(log(y)) (By definition of the density fn "f" and the fundamental thm of calculus)
So FX(x) = integral from negative infinity to log(y) of fX(x) because of more by definition crap. I have no intention of typing the density function again.
Aaaaand I just have noticed that I the solution I got was potentially correct because I was reading the wrong solution. There just might be a log^2(y) in there.
I will burn this book.
Hyolith, none of my books use "ln." None of my professors use "ln." If you type "log(e)" into a mathematical program, it assumes the natural log and gives you one. I haven't seen a log base ten in several years.
As a matter of convention, "log" is generally assumed to be the natural log because log base ten is pretty much worthless.
Edited, May 5th 2011 12:10pm by Sweetums
As a matter of convention, "log" is generally assumed to be the natural log because log base ten is pretty much worthless.
Edited, May 5th 2011 12:10pm by Sweetums
Sweetums wrote:
Hyolith, none of my books use "ln." None of my professors use "ln." If you type "log(e)" into a mathematical program, it assumes the natural log and gives you one. I haven't seen a log base ten in several years.
As a matter of convention, "log" is generally assumed to be the natural log because log base ten is pretty much worthless.
Edited, May 5th 2011 12:10pm by Sweetums
As a matter of convention, "log" is generally assumed to be the natural log because log base ten is pretty much worthless.
Edited, May 5th 2011 12:10pm by Sweetums
Ah, then it's just a difference in how we were taught. In my classes we have always used ln() to express the natural log, and really haven't dealt with log base # in a long time.
Sweetums wrote:
Deadgye wrote:
Wiki tells me normal distribution is (1/sqrt(2pi)) * e^(-.5x^2). It's either you need to integrate that, or you need to replace x with that in y=e^x and integrate that, or you need to set x = (1/sqrt(2pi)) * e^(-.5z^2) and double integrate y = e^x with respect to z first, then x.
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
The density function fX(x) is given but you need to find fY(y), so you pretty much take the inverse of Y=eX so you can put the distribution function in terms of X.
FY(y) = Pr(Y≤y) which is equivalent to Pr(eX≤y) and then you take the log so you get FY(y) = FX(log(y))
d/dx (FX(log(y))) = fX(log(y)) (By definition of the density fn "f" and the fundamental thm of calculus)
So FX(x) = integral from negative infinity to log(y) of fX(x) because of more by definition crap. I have no intention of typing the density function again.
Aaaaand I just have noticed that I the solution I got was potentially correct because I was reading the wrong solution. There just might be a log^2(y) in there.
I will burn this book.
Is Fx(x) what you are trying to find then? If so I was kind of close when I said
Hyolith wrote:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
Except that Y should be the log(x) and f(y) is that big *** long function and it would be in terms of x and not y.
I would plug it into my TI-89 if I hadn't checked it last night and noticed the batteries have completely corroded. [:sad:]
They're essentially interchangeable. If you want to denote any other logarithm you pretty much have to use logn.
Hyolith wrote:
Sweetums wrote:
Deadgye wrote:
Wiki tells me normal distribution is (1/sqrt(2pi)) * e^(-.5x^2). It's either you need to integrate that, or you need to replace x with that in y=e^x and integrate that, or you need to set x = (1/sqrt(2pi)) * e^(-.5z^2) and double integrate y = e^x with respect to z first, then x.
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
Whatever you're taking is very similar to the class I'm taking this semester, but unfortunately I've fallen behind and I'm not catching up until after next week. (Binge studying before the final weee)
Edited, May 5th 2011 12:00pm by Deadgye
The density function fX(x) is given but you need to find fY(y), so you pretty much take the inverse of Y=eX so you can put the distribution function in terms of X.
FY(y) = Pr(Y≤y) which is equivalent to Pr(eX≤y) and then you take the log so you get FY(y) = FX(log(y))
d/dx (FX(log(y))) = fX(log(y)) (By definition of the density fn "f" and the fundamental thm of calculus)
So FX(x) = integral from negative infinity to log(y) of fX(x) because of more by definition crap. I have no intention of typing the density function again.
Aaaaand I just have noticed that I the solution I got was potentially correct because I was reading the wrong solution. There just might be a log^2(y) in there.
I will burn this book.
Is Fx(x) what you are trying to find then? If so I was kind of close when I said
Hyolith wrote:
E[Y] = integral from -inf to +inf of Y*f(Y)dy where f(y) would be like f(x) above only with "y."
Except that Y should be the log(x) and f(y) is that big *** long function and it would be in terms of x and not y.
I would plug it into my TI-89 if I hadn't checked it last night and noticed the batteries have completely corroded. [:sad:]
I'm trying to find a good PDF on google to explain it in case you wanna read about it, since you seem interested. If anything, try "transform of random variables probability" and you'll probably find somethin'.
Edited, May 5th 2011 12:40pm by Sweetums
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