Math  340                    Homework  Number 14 
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1.   (#3,  page 309)

U is uniform on (0, 1)  =>  U has the following desity

f_{U} (u)  =   1        for  0 < u < 1
                  =    0       otherwise

Now,  we are interested in U^2,  namely a change of variable as follows:

V  =  g(U)  =  U^2

Thus,  the new r.v.  V  has a density  f_{V} (v)  given by:

f_{V}(v)  =  f_{U}(u)  /  g^{prime}(u)

                 =  f_{U} (u)  /  2 u

                 =  f_{U} (sqrt(v))  /  2 sqrt(v)       since  v = u^2

Note that as  u  varies from  0  to  1,  v will do the same.  Hence,  we
may conclude that

f_{V} (v)   =  1 / 2 sqrt(v)      for  0 < v < 1
                   =   0                         otherwise
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2.   (#5,  page 309)

X  is uniform on [-1,  2].   Thus it has the following density:

f_{X} (x)   =   1/3      for  -1 \leq  x  \leq  2
                   =    0          otherwise

Y  =  g(X)  =  X^2

=>     f_{Y} (y)  =  f_{X} (x)  /  g^{prime} (x)

                           =  f_{X} (x)  /  2 x

                           =  f_{X} (sqrt(y))  /  2 sqrt(y)

Note that as  x  varies from -1  to  2,  y varies from 1 to 4.  Also note
that as x varies from -1 to 0  y attains the same values as when x
varies from 0 to 1.  So,  for this will result in a doubling up of the
probabilities for y as it varies from 0 to 1.  Hence,

f_{Y} (y)  =  1 / 3 sqrt(y),   for  0 < y < 1

                  =  1 / 6 sqrt(y),   for  1 < y < 4.
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3.   (#3,  page 323)

a.  By symmetry  Y has the same distribution as  X.

b.  The event  (R \leq  r}  is the set of all points in the unit disk whose
distance from the origin is less than or equal to r,  namely,  it is the
disk of radius r together with its boundry.

In order to DEDUCE a formula for the c.d.f. of R we consider the
following:

i.  F_{R}(r)  =  integral_{0}^{r}  f_{R}(r)  dr

ii.  As  r -> 1,  F_{R}(r) -> 1.

iii.  F_{R}(0)  =  0

iv.  f_{R}(r)  =  P(R \in dr)  =  2 pi r dr \phi(x) \phi(y)

So,  we deduce that it should be of the form r^{2}.
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4.   (#5,  page 323)

Recall that the cumulative distribution function,  F(x),  associated with
a density  f(x)  is  defined as:

F(x)  :=  P( X  \leq  x)  =  integral_{-infinity}^{x}   f(t) dt

for each  x in  (-infinity,  infinity).  Now,  in order to compute the
cumulative dist func for the given density we note that we need to
consider two cases  x \leq 0  and  x \geq 0  due to the presence of the
absolute value sign.  

Case  1:     x  \geq  0.

F(x)  =  int_{ - infty}^{x}  1/2  e^{- |t|}  dt

          =  1/2   +  1/2  int_{0}^{x} e^{- t}  dt

          =  1/2  -  1/2  e^{-t} |_{0}^{x}

          =  1/2  -  1/2 (e^{-x}  -  1)

          =  1  -  1/2 e^{-x}

using the fact that the integral from minus infinity to zero is equal to
half of the total area under the density function,  which makes it equal
to 1/2.


Case  2:  x  \leq  0.

F(x)  =  int_{ - infty}^{x}  1/2  e^{t}  dt

          =  1/2   e^{t}   |_{-infty}^{x}

          =  1/2  e^{x}  -  0

          =  1/2  e^{x}