ECE353 Midterm 2 solution
Problem1 A limiter is shown in figure below. Assuming X is random variable with uniform
distribution between -2 and 2
Y=g(X)
1
-2
X
-1
1
2
-1
Find E{Y2}.
Solution
 1 ,2  X  1

Since Y  g ( X )   X ,1  X  1
1
,1  X  2

1
2
E{Y 2 }  E{g ( X ) 2 } 
2
2
 g ( x) f X ( x)dx   (1)
2
2
1
1
x 
x
x
     
 4  2 12  1  4 1
1 2 1
  
4 12 4
2

3
3
2
1
2
1
1
1
dx   x 2
dx   (1) 2
dx
(2  (2))
(2

(

2))
(2

(

2))
1
1
Problem2
Let Y  e X . If X is exponentially distributed with parameter   1.
a) What is the range of Y
Solution
Since X is an exponentially distributed random variable, then X  x, where 0  x   .
Therefore, the range of Y is Y  y, where 1  y   .
b) Determine and sketch f Y ( y) ?
Solution
d
 g 1 ( y) 
dy
d
d
1
, where x  g 1 ( y)  ln y . Thus,
g 1 ( y )   ln( y ) 

dy
dy
y
fY ( y )  f X ( g 1 ( y ))*
fY ( y )  1 e1ln( y ) *
1
1
1 1
 ln( y ) *  2 , for 1  y  
y e
y y
1.5
f Y(y)
1
0.5
0
1
5
10
15
y
20
25
Problem3
Random variables X and Y have pdf
c x  y  1, x  0, y  0
f ( x, y )  
0 otherwise
a) Sketch the region of nonzero probability
Y
x +y = 1
1
f (x,y)>0
0
1
b) What is the value of constant c
Solution
Let

f XY ( x, y )dxdy  1
R:( x , y ), f ( x , y )  0
1 1 y
  cdxdy  1
0 0
1
 c(1  y )dy  1
0
c
(c  )  1
2
c2
X
c) What is value of P(Y  X  0.5) ?
Y
x +y = 1
y = x +0.5
1
0
0.25
X
1
Solution
P{Y  X  0.5}

=
f XY ( x, y )dxdy  1
R:( x , y ), P{Y  X  0.5}
0.25 1 x
=
 
2dydx
0 x  0.5
0.25
=
2

0.25
(1  x)  ( x  0.5)dx =
0
=
(2 x 2  x)
=
1
8
 (4 x  1)dx
0
0.25
0
This can also be solved by noticing that the area is 1/8 of the entire area. Since it is uniformly
distributed over the entire region, we can immediately conclude that P{Y  X  0.5} =1/8.
Bonus Problem
Suppose you arrive at a bus stop at time 0, and at the end of each minute, with probability
p, a bus arrives, or with probability 1-p, no bus arrives. Whenever a bus arrives, you board that
bus with probability q and depart. Let T equal the number of minutes that you stand at a bus stop.
Let N be the number of buses that arrive while you wait at the bus stop.
a) Identify the set of points (n, t) for which P(N=n, T=t) > 0.
Solution
{(n,t)| t ≥ n where, t > 0, and n > 0}
b) Find PNT(n, t).
Solution
P{buses arrives at T=t| boarding the bus}P{boarding the bus}
 t  1  n 1
t n
n 1
Thus, PNT (n, t )  
 p (1  p) *(1  q) * p * q .
 n  1