Solution: we first find the cdf and then the pdf as follows:
if y>0. For y<0 note that P(-logX<y) = 0 because 0<X<1, so logX<0, so -logX>0 always.
Example say X~U[0,1] and l>0. What is the pdf (pmf?) of the random variable Y=-llog(X)?
Solution: we first find the cdf and then the pdf as follows:
if y>0. For y<0 note that P(-logX<y) = 0 because 0<X<1, so logX<0, so -logX>0 always.
An R check of this result is in the R function gen_exp
This is an example of a function (or transformation) of a random variable. These transformations play a major role in probability and statistics. We will see how to find their pdf's (pmf's) on a few examples.
Example say X is the number of roles of a fair die until the first six. We have already seen that P(X=x) = 1/6*(5/6)x-1, x=1,2,.. Let Y be 1 if X is even, 0 otherwise. Find the pmf of Y.
Note: here both X and Y are discrete.
Solution:
and P(Y=0) = 1 - P(Y=1) = 5/11
R test:
x=rgeom(10000,1/6)+1
y=1-x%%2
table(y)/10000
Example say X is a continuous r.v with pdf fX(x) = ½exp(-|x|) x
(this is called a double exponential) Let Y=I[-1,1](X). Find the pmf of Y.
Note: here X is continuous and Y is discrete.
R check: gen_dexp(1)
Example again let X have pdf fX(x) = ½exp(-|x|) x . Let Y =X2. Then for y<0 we have P(Y≤y) = 0. So let y>0. Then
R check: gen_dexp(2)
Next up some examples of functions of random vectors:
Example Say (X,Y) is a discrete rv with joint pmf fX,Y(x,y)=(1-p)2px, x,y{0,1,..}, y≤x, and 0<p<1. Let U=X and V=X-Y. Find fU,V(u,v)
fU,V(u,v) = P(U=u,V=V) = P(X=u,X-Y=v) = P(X=u,u-Y=v) = P(X=u,Y=u-v) = (1-p)2pu
where u,v{0,1,2..} and v=x-y≤x=v. So we see that fU,V(u,v)=fX,Y(u,v), or
(X,Y) has the same distribution as (U,V)!
Example say (X,Y) is a bivariate standard normal r.v, that is it has joint density given by
for (x,y) 2
Let the r.v. (U,V) be defined by U = X+Y and V = X-Y. Find the joint pdf of (U,V).
To start let's define the functions g1(x,y) = x+y and g2(x,y) = x-y, so that U=g1(X,Y) and V = g2(X,Y).
For what values of u and v is f((U,V)(u,v) positive? Well, for any values for which the system of 2 linear equations in two unknowns u=x+y and u=x-y has a solution. These solutions are
x = h1(u,v) = (u + v)/2
y = h2(u,v) = (u - v)/2
From this we find that for any (u,v) 2 there is a unique (x,y) 2 such that u=x+y and v=x-y. So the transformation (x,y) → (u,v) is one-to-one and therefore has a Jacobian given by
Now from multivariable calculus we have the following:
Note that the density factors into a function of u and a function of v. This is not only a necessary but also a sufficient condition for U and V to be independent.
Example A rv X is called a normal (or Gaussian) rv with mean m and standard deviation s if it had density
a special case is a standard normal rv, which has m=0 and s=1
Say X and Y are independent standard normal rv's. Let Z = X + Y. Find the pdf of Z.
Note: now we have a transformation from 2 → .
Let's first assume that X and Y are standard normal rv's, that is mX=mY=0 and sX=sY=1. Then Z = X + Y = U in the example above, so the pdf of Z is just the marginal of U and we find
Say X and Y are two continuous independent r.v with pdf f's fX and fY, and let Z = X+Y. If we repeat the above calculations we can show that in general the pdf of Z is given by
This is called the convolution formula.
There is a second method for deriving the convolution formula which is useful, using the law of total probability for rv's:
Example say X and Y are independent exponential rv's with rate l. Find the pdf of Z=X+Y
R: gen_sumexp(lambda)
Example Say (X,Y) is a discrete rv with joint pmf fX,Y(x,y)=(1-p)2px, x,y{0,1,..}, y≤x, and 0<p<1. Let U=I(X=Y). Find fU(u)
Example : Say X1, .., Xn are iid U[0,1]. Let M=max{X1, .., Xn}. EM and VM.
First we find fM
Now
R: gen_maxunif(n)