8614smith
Nov18-10, 10:47 AM
here is the background information before the question:
For a function of x, y(x), the two probability distributions must satisfy \left|p_{y}(y)dy\right|=\left|p_{x}(x)dx\right| and therefore
p_{y}(y)=p_{x}(x)\left|\frac{dx}{dy}\right| where \int^{y_{max}}_{y_{min}}p_{y}(y)dy=\int^{x_{max}}_ {x_{min}}p_{x}(x)dx=1
Now, if we want to generate a particular probability distribution, p_{y}(y)=f(y), and for simplicity let us assume that y(x) is a monotonically increasing function of x, then since x is a uniform deviate this implies
1x=\int^{y(x)}_{y(0)}f(y)dy=F[y(x)] \frac{dF}{dy}=f(y) F[y(0)]=0 F[y(1)]=1
so that,
y(x)=F^{-1}(x) y(x_{max})=y(1)=F^{-1}(1)=y_{max} F[y(x_{max})]=F(y_{max})=x_{max}=1
Given a uniform deviate x (as described above) find y(x) such that the distribution function for y, p_{y}(y), will be equal to f(y), and find the value of either y_{max} or the normalization constant A for which p_{y}(y) will be properly normalized in the range y_{min}<y<y_{max}.
(a) f(y)=Ay+1 y_{min}=0 y_{max}=2; find A and y(x)
(b)f(y)=2y+4y^{3} y_{min}=0; find y_{max} and y(x)
Attempt at the answer;
(a)
using 1 i have x=\frac{Ay(x)^{2}}{2}+y(x)
substituting for x_{max}, which i guessed must be 1? as its the max probability which is unity.
1=\frac{Ay(x)^{2}}{2}+y(x)
putting in the limits of y_{min}=0 and y_{max}=2 into the integral gives;
-1=2A\Rightarrow{A=-0.5}
is that correct? and how do i find y(x)??
(b)
using 1 and substituting the lower limit of y_{min}=0 i have x=y^{2}+y^{4}
how do i go to find y_{max} from here? i can put x = 1, but then i get stuck again as i can't get y(x) on its own to get a value for it.
For a function of x, y(x), the two probability distributions must satisfy \left|p_{y}(y)dy\right|=\left|p_{x}(x)dx\right| and therefore
p_{y}(y)=p_{x}(x)\left|\frac{dx}{dy}\right| where \int^{y_{max}}_{y_{min}}p_{y}(y)dy=\int^{x_{max}}_ {x_{min}}p_{x}(x)dx=1
Now, if we want to generate a particular probability distribution, p_{y}(y)=f(y), and for simplicity let us assume that y(x) is a monotonically increasing function of x, then since x is a uniform deviate this implies
1x=\int^{y(x)}_{y(0)}f(y)dy=F[y(x)] \frac{dF}{dy}=f(y) F[y(0)]=0 F[y(1)]=1
so that,
y(x)=F^{-1}(x) y(x_{max})=y(1)=F^{-1}(1)=y_{max} F[y(x_{max})]=F(y_{max})=x_{max}=1
Given a uniform deviate x (as described above) find y(x) such that the distribution function for y, p_{y}(y), will be equal to f(y), and find the value of either y_{max} or the normalization constant A for which p_{y}(y) will be properly normalized in the range y_{min}<y<y_{max}.
(a) f(y)=Ay+1 y_{min}=0 y_{max}=2; find A and y(x)
(b)f(y)=2y+4y^{3} y_{min}=0; find y_{max} and y(x)
Attempt at the answer;
(a)
using 1 i have x=\frac{Ay(x)^{2}}{2}+y(x)
substituting for x_{max}, which i guessed must be 1? as its the max probability which is unity.
1=\frac{Ay(x)^{2}}{2}+y(x)
putting in the limits of y_{min}=0 and y_{max}=2 into the integral gives;
-1=2A\Rightarrow{A=-0.5}
is that correct? and how do i find y(x)??
(b)
using 1 and substituting the lower limit of y_{min}=0 i have x=y^{2}+y^{4}
how do i go to find y_{max} from here? i can put x = 1, but then i get stuck again as i can't get y(x) on its own to get a value for it.