Transforming a uniform distribution into a binomial

[bennyska]

Homework Statement

Let X~UNIF(0,1). Find y = G(u) such that Y = G(U)~BIN(3,1/2)

Homework Equations

The Attempt at a Solution

after a bit of searching/reading, i found how to do this with a continuous distribution (the problem i had was an exponential, so i took the inverse)... however, more searching has not led to any results for the discrete case, which i need for the binomial. any pointers would be definitely appreciated.

 

[micromass]

You might want to take a look at http://en.wikipedia.org/wiki/Inverse_transform_sampling

Basically, given X an arbitrary distribution, and let F_X be it's cdf, then we define

$$F^{-1}_X(u)=inf\{x~\vert~F(x)=u,~0<u<1\}$$.

Then, if U has uniform distribution, then $F_X^{-1}(U)$ has the distribution of X...

 

[vela]

X is continuous whereas Y is not. This implies that G(X) isn't be continuous. Does that help?

 

[bennyska]

X is continuous whereas Y is not. This implies that G(X) isn't be continuous. Does that help?

no... how about one more hint?

 

[micromass]

What do you mean, one more hint? I practically gave you the answer in my post!

 

[vela]

Start by figuring out the cdf for Y.

 

[bennyska]

well, for a binomial, with these parameters, the cdf would be .$$.5^{3}\sum_{i=0}^{3}\left({3\atop i}\right)$$

 

[vela]

That expression is not the cdf. That's equal to 1. Try again.

 

[bennyska]

$$F_{x}=Pr(X\leq x)=.5^{3}\sum_{i=1}^{x}\left({x\atop i}\right) for<br /> x = 0, 1, 2, 3[\tex]<br /> <br /> not sure if this tex will show up, but here's the pdf<br /> <br /> (also, basically .5³*SUM(from i=0 to x) (x choose i) )$$

 

[vela]

Close enough. The lower limit of the summation should be i=0, but other than that it's correct. It's easy enough to just write F_X(x) out explicitly.
$$F_X(x) = \left\{\begin{array}{cl} <br /> 0 & \textrm{if}~x < 0 \\<br /> 1/8 & \textrm{if}~~0 \le x \lt 1 \\<br /> 1/2 & \textrm{if}~~1 \le x \lt 2 \\<br /> 7/8 & \textrm{if}~~2 \le x \lt 3 \\<br /> 1 & \textrm{if}~~3 \le x <br /> \end{array}\right.$$
Does this give you an idea of how you might write Y=G(X)?

 

[vela]

That should have been F_Y(y), not F_X(x).

 

[bennyska]

$$G_{X}(y)=\begin{cases}<br /> 0 & 0\leq x<1/8\\<br /> 1 & 1/8\leq x<1/2\\<br /> 2 & 1/2\leq x<7/8\\<br /> 3 & 7/8\leq x\end{cases}$$
?

 

[I like Serena]

Cheers!