Post by: Moko on August 26, 2012, 01:06:20 pm
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
Post by: FlorianK on August 26, 2012, 02:45:10 pm
Ok this question is really bugging me:
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
I didn't thought longer than 10sec for that so I'm probably wrong cause it's to easy.
But wouldn't it be just 3(x/3)2/2
Post by: Moko on August 26, 2012, 07:40:03 pm
they've got:
x2/18 if -3 <(or equal to) x <(or equal to) 3 and 0 otherwise
If you can get it, please show me how
Post by: TrueTears on August 26, 2012, 07:54:35 pm
No! However good try! Your answer may be correct, however the method you used is flawed. The reason for this is actually pretty complicated (you need'nt worry about this in VCE, it is MUCH beyond VCE), in general, if X has a probability density function of f(x) then you cannot conclude that g(X) has a PDF of f(g(x)). There is a whole branch of probability theory that focuses on finding out the new probability distribution of transformations of random variables.Ok this question is really bugging me:
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
I didn't thought longer than 10sec for that so I'm probably wrong cause it's to easy.
But wouldn't it be just 3(x/3)2/2
Just a teaser of what they are:
Moment generating functions: http://en.wikipedia.org/wiki/Moment-generating_function
Cumulant generating functions: http://en.wikipedia.org/wiki/Cumulant
Method of transforms
Method of distributions
For this particular question, I don't why the hell they would ask it in methods but it is infact a very deep question that requires quite a bit of fundamental knowledge of probability theory (as I don't recall them teaching you guys what a distribution function is in VCE lol).
To do this question, there are 2 methods we can use: Either the Method of Distribution Functions or the Method of Transformations: (I will not show the MGF and CGF approaches)
(http://img571.imageshack.us/img571/3622/asdfasdfd.jpg)
I will show both methods:
First using the Method of Distribution Functions:
Let
We then have
Now
Now since
As required.
We can get the same answer by applying the method of transforms.
Again let Y = 3X
I will mechanically apply the formula (there is a good reason why the formula is what it is, however to explain would be well beyond VCE, the formula itself, is well within VCE boundaries though)
First we solve for X, yielding
Then
To get the domain, it is the same as the previous method.
For reference, if you wish to read up on these methods, since I actually cannot find any good wiki articles on them, so you can have a read of Mathematical Statistics by Wacklery, chapter 6 :)
Post by: TrueTears on August 26, 2012, 08:07:53 pm
yeah man you're right...it is too easy! lolAlso they really should have used another variable, say Y or Q or anything, by using X again is very bad notation since it is the original PDF and not the transformed PDF
they've got:
x2/18 if -3 <(or equal to) x <(or equal to) 3 and 0 otherwise
If you can get it, please show me how
Post by: brightsky on August 26, 2012, 09:31:58 pm
Ok this question is really bugging me:
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
crude, albeit intuitive, explanation:
imagine a probability distribution table with x-values at the top and the corresponding probabilities at the bottom. 3X would mean multiply 3 to the x-values. so essentially you are dilating the pdf by a factor of 3 from the y-axis. so for now, the rule is: y = 3(x/3)^2/2 = x^2/6, where x E [-3,3]. but we have a problem! if we do integrate that function from -3 to 3, we get 3 instead of 1! so we need to tweak the function slightly so that it remains a pdf. in this case, we need to multiply it by 1/3. so the final function is y = x^2/6 * 1/3 = x^2/18.
the above intuition may help you undestand the relationship between normal distribution and standard normal distribution. the pdf for standard normal distribution is f(x) = 1/sqrt(2*pi)*e^(-1/2 x^2). now how to obtain the pdf for normal distribution from that? well we need to dilate by factor of o from the y-axis and then translate u units to the right. but if we calculate the area under the graph now, we find that it becomes o units^2, instead of 1. so we need to divide by o to account for this change in area. thus the pdf for normal distribution is y = 1/(o*sqrt(2pi)) * e^(-1/2*((x-u)/o)^2). note well the o in the denominator.
Post by: Moko on August 26, 2012, 09:36:56 pm
Thanks a lot man!
Post by: TrueTears on August 26, 2012, 09:37:25 pm
Very nice! That's a very really good intuitive way of thinking about it hahaOk this question is really bugging me:
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
crude, albeit intuitive, explanation:
imagine a probability distribution table with x-values at the top and the corresponding probabilities at the bottom. 3X would mean multiply 3 to the x-values. so essentially you are dilating the pdf by a factor of 3 from the y-axis. so for now, the rule is: y = 3(x/3)^2/2 = x^2/6, where x E [-3,3]. but we have a problem! if we do integrate that function from -3 to 3, we get 3 instead of 1! so we need to tweak the function slightly so that it remains a pdf. in this case, we need to multiply it by 1/3. so the final function is y = x^2/6 * 1/3 = x^2/18.
the above intuition may help you undestand the relationship between normal distribution and standard normal distribution. the pdf for standard normal distribution is f(x) = 1/sqrt(2*pi)*e^(-1/2 x^2). now how to obtain the pdf for normal distribution from that? well we need to dilate by factor of o from the y-axis and then translate u units to the right. but if we calculate the area under the graph now, we find that it becomes o units^2, instead of 1. so we need to divide by o to account for this change in area. thus the pdf for normal distribution is y = 1/(o*sqrt(2pi)) * e^(-1/2*((x-u)/o)^2). note well the o in the denominator.
Post by: FlorianK on August 26, 2012, 10:04:18 pm
Yeah forgot about the integration part.Ok this question is really bugging me:
Let X be a random variable with probability density function:
f(x) =3x2/2 -1<(or equal to) x <(or equal to) 1
Find the rule of a probability density function for 3X.
crude, albeit intuitive, explanation:
imagine a probability distribution table with x-values at the top and the corresponding probabilities at the bottom. 3X would mean multiply 3 to the x-values. so essentially you are dilating the pdf by a factor of 3 from the y-axis. so for now, the rule is: y = 3(x/3)^2/2 = x^2/6, where x E [-3,3]. but we have a problem! if we do integrate that function from -3 to 3, we get 3 instead of 1! so we need to tweak the function slightly so that it remains a pdf. in this case, we need to multiply it by 1/3. so the final function is y = x^2/6 * 1/3 = x^2/18.
the above intuition may help you undestand the relationship between normal distribution and standard normal distribution. the pdf for standard normal distribution is f(x) = 1/sqrt(2*pi)*e^(-1/2 x^2). now how to obtain the pdf for normal distribution from that? well we need to dilate by factor of o from the y-axis and then translate u units to the right. but if we calculate the area under the graph now, we find that it becomes o units^2, instead of 1. so we need to divide by o to account for this change in area. thus the pdf for normal distribution is y = 1/(o*sqrt(2pi)) * e^(-1/2*((x-u)/o)^2). note well the o in the denominator.
How do you get the divided by 2 for the dilation?
For me it would just be:
g(x) is the pdf for 3X
dilation by factor 3.
f(x)=(x/3)²
g(x)=a(x/3)²
F(x)=(x/3)³
integral of f(x) from -3 to 3 = 1+1=2
therefor a is equal to 0.5
g(x)=0.5(x/3)²=x²/18