Download Homework 7

CMPE 107 - Homework 7
Solutions
NOTE: If a RV is uniform with parameters a and c, it means that it is uniformly
distributed over the interval [a-c/2,a+c/2]
1. The voltage x across a 1 Ω resistor is a uniform random variable with a = 0.5 and
c = 1. The instantaneous power is y=x2. Find the cdf Fy(y) and the pdf fy(y) of y.
Note that Fx(x) = x for 0≤x≤1< =0 for x<0, and =1 for x>1.
For y0 < =0 : Iy0 = {x: x2<y0} is the empty set, so Fy(y0) = 0
# y y0 " 1
For y0 > 0 : Iy 0 = [! y 0 , y0 ] , therefore Fy ( y0 ) = Fx ( y0 ) ! Fx ( ! y 0 ) = $ 0
% 1 y0 > 1
Note that Fy(0)=0.
1
The pdf fy(y) is the derivative of the cdf, therefore it is 0 for y≤0,
for 0<y≤1, and 0
2 y
for y>1 (note the discontinuity points at 0 and 1).
2. x is an exponential random variable with parameter c. What is the cdf of y = x ?
Remember that the cdf of an exponential RV is (1 ! e !cx )U( x)
If y<0, Iy is the empty set, so Fy(y)=0
!cy 2
If y>0, Iy = ( x : x ! y) = [0, y2 ] , therefore Fx(y) = Fx(y2)-Fx(0) = 1 ! e
Notice that Fy(0)=0.
3. x is a uniform random variable with a=1 and c=2. The random variable w is the
"x x ! 1
output of the clipper: w = g(x ) = #
Find the cdf Fw(w) and the pdf fw(w).
$1 x > 1
Note that the cdf of x is x/2 for 0≤x≤2.
For w<1, Iw = ( x : g(x) ! w) = ["#,w] , and therefore Fw(w) = Fx(w) (so fw(w)=fx(x)).
For w>1, Fw(w) = 1 (so fw(w)=0). For w=1, there is a discontinuity in Fw(w) (because w
takes value 1 with probability 0.5). Hence, fw(w) is not defined there (or better, it is equal
to 0.5 δ(w-1).
4. A dealer’s profit, in units of $5,000, on a new automobile is given by y=x2, where
"2(1 ! x ), 0 < x < 1
x is a random variable having the density function f (x) = #
elsewhere
$ 0,
•
Find the pdf of y
•
Find the probability that the profit will be less than $500 on the next new
automobile sold by this dealership.
5. Let x be a binomial random variable with n=3 and p=2/5. Find the probability
distribution of the random variable y=x2.
! 3$ ! 2 k ! 3 3&k
The distribution of x is f x ( k ) = # " $% " $% for k={0,1,2,3}. Therefore,
" k% 5 5
m
3& m
! 3 $ ! 2$ ! 3$
f y ( m) = #
for m={0,1,4,9}.
" m% " 5 % " 5 %
6. Let y=x2 where x is an exponential random variable with parameter c = 9. What is
P(y>1)?
Remember that the cdf of an exponential RV is (1 ! e !cx )U( x)
P(y>1) = 1-P(y≤1) = 1-P(-1≤x≤1) = 1 ! (1! e ! c ) = e ! c
7. The voltage V at the output of a microphone is a uniform random variable with
limits –1 volt and 1 volt. The microphone voltage is processed by a hard limiter
with cutoff points –0,5 volt and 0.5 volt. The magnitude of the limiter output L is
$ V , V " 0.5
&
a random variable such that L = % 0.5 , V > 0.5
&#0.5 , V < -0.5
'
 What is P(L=0.5)?
 What is FL(L)?
!
The cdf of L is equal to 0 for L≤-0.5, it is equal to FV(L) for
-0.5≤L≤0.5, and it is equal to 1 for L>0.5. Remember that here, FV(V)=(V+1)/2 for –
1≤V≤1, =0 for V<1, and =1 for V>1. P(L=0.5) = 1 – (0.5+1)/2 = 0.25 (this is the
discontinuity of FV(V) at 0.5).
8. The speed of a molecule in a uniform gas at equilibrium is a random variable v
"kv2 e ! bv 2 ,
v>0
whose density is f v (v ) = #
where k is an appropriate constant
elsewhere
$ 0,
and b depends on the absolute temperature and mass of the molecule. Find the pdf
of the kinetik energy w of the molecule, where w=mv2/2.
For w>0, the function w=g(v)=mv2/2 has two roots: v = ±
roots. The derivative of g(v) at point v = ±
2w
. For w<0, there are no
m
2w
is mv = ± 2mw . Therefore, the solution
m
is
" 0 + 2kwe !2bw / m 2kwe !2bw / m
=
,
w>0
f w( w) = # m 2mw
m 2mw
0,
elsewhere
$
!2x,
9. Given the random variable f (x) = "
# 0,
3
y=8x .
The root of the function y=g(x)=8x3 is x=
3
0 < x <1
elsewhere
find the pdf of y, where
3 y
y
. The derivative of g(x) at point
is 6y2/3.
2
2
Therefore,
3 y
"$ 3 y
y 1/ 3 y !2 / 3
1
" 1 , 0< y<8
=
= 3 , 0<
<1
2/
3
f y ( y ) = # 6y
= # 63 y
6
6 y
2
$%
elsewhere
0,
elsewhere % 0,
As a sanity check, you can verify that fy(y) integrates to 1.