Random Vectors
Shruti Sharma
Ganesh Oka
References :Probability and Random Processes for Electrical Engineering: Leon-Garcia
Applied Stochastic Processes: Lefebvre, M
Random Vectors

“An n- dimensional random vector is a
function X = (X1….,Xn) that associates a vector
of real numbers with each element s of a
sample space S of a random experiment E.”

Xk is a vector of random variables

Sk is a set of all possible values of X
Examples of Random Vectors

Discrete r.v. :- A semiconductor chip is divided into
‘M’ regions. For the random experiment of finding the
number of defects and their locations, let Ni denote
the number of defects in ith region. Then
N() = (N1(),…,NM()) is a discrete r.v.
Continuous r.v. :- In a random experiment of selecting
a student’s name, let
H() = height of student  in inches.
W() = weight of student  in pounds.
A() = age of student  in years.
Then (H(), W(), A()) is a continuous r.v.

Product form events.
For the n-dimensional r.v. X = (X1,…,Xn), a product form
event A can be expressed as follows.

A = {X1 in A1}  {X2 in A2}  …  {Xn in An}, where Ak is a
1-dimensional event involving only Xk.
This helps when the random vectors are dependent.
 Not all events can be can be expressed in product form.

Examples
A = {X+Y  10}
B = {min(X,Y)  5}
C = {X2+Y2  100}
Example.
Let X be the input to a communication channel and let Y be
the output. Input is +1 or -1 volt with equal probability. Output
is the input plus a noise voltage that is uniformly distributed in
the interval from -2 to +2 volts. Find the probability of positive
input but not positive output.

To find
P[X = +1, Y  0] = P[ {X = +1}  { Y  0} ] = P[ { Y  0 }  { X = +1 } ]
= P[ Y  0 | X = +1] P[ X = +1]
P[ X = +1 ] = ½.
When X = +1, Y is uniformly distributed in the interval [-2+1, 2+1] = [-1, 3]
Now, P[ Y  y | Y  [-1, 3] ] = (y – (-1)) / ( 3 – (-1))
Thus,
P[ Y  0 | X = +1 ] = P[ Y  0 | Y  [-1, 3] ] = ¼.
 P[ X = +1, Y  0 ] = ¼ . ½ = 1/8
Two Dimensional Random Vector

Random Vector (r.v) Z = (X,Y), where
X = xj, j = 1,2,… Y = yk, k = 1,2,…
Or X and Y are continuous.

Joint Distribution Function:
FX,Y(x,y) = P[{X<x} ∩ {Y<y}] = P[X<x,Y<y]

Marginal Distribution Function:
FX(x) = P[X<x, Y<∞] = FX,Y(x, ∞)
FY(y) = P[X<∞, Y  y] = FX,Y(∞, y)
And  FX =  FY = 1 --- discrete case
Or  FX =  FY = 1 --- continuous case
Properties

FX,Y(-∞ ,y) = FX,Y(x,-∞) = 0

FX,Y(∞ , ∞) = 1 ----- Normalization condition.

FX,Y(x1 , y1) < FX,Y(x2 , y2) if x1<x2 and y1< y2

P[a<X<b, c<Y<d] = FX,Y(b,d) - FX,Y(b,c) FX,Y(a,d) + FX,Y(a,c)
a, b, c and d are constants
Discrete Type Random Vectors



A two dimensional discrete type r.v. has a SZ
that’s finite, or countably infinite.
SZ = SX * Y = {(xj ,yk), j = 1,2….. k = 1,2……}
Joint Probability Mass Function
pX,Y(xj,yk) = P[X=xj, Y = yk]
Marginal Probability Mass Functions
pX(xj) = Σ pX,Y(xj,yk) and pY(yk) = Σ pX,Y(xj,yk)
all yk
all xj
Continuous Random Vector


A two-dimensional r.v. Z= (X,Y) is continuous
if SZ is a uncountably infinite subset of R2.
Joint Probability Density Function
fX,Y(x,y) = 2 FX,Y(x,y)
xy
Marginal
Probability Density Function

fX(x) = ∫ fX,Y(x,y)dy
-
Probability of event Z that belongs to A:
∫ ∫ fX,Y (x,y) dxdy

A
Distribution Functions

For the Discrete case:
FX,Y (x,y) = Σ pX,Y(xj ,yk)
xjx,yky
For the Continuous Case:
x
F X ,Y ( x , y ) 
y


f X ,Y ( u , v ) dvdu
Example --- marginal pdf

fX,Y(x,y) = (ln x / x)
= 0

f X ( x) 

x
f X ,Y ( x , y ) dy 


e
f X ,Y ( x , y ) dx 



1
e
fY ( y ) 

f X ,Y ( x , y ) dy  ln x
-----If 1  x  e
0

fY ( y ) 
if 1< x< e, 0< y<x
otherwise
f X ,Y ( x , y )dx 
ln x
dx 
x
2
1  (ln y )
y
And fY(y) = 0 otherwise.
1
2
-----If 0 < y < 1 ( x)
2
-----If 1  y (< x) < e
Independent Random Variables

If (X,Y) is a random vector, X,Y are
independent variables if :




pX,Y(xj,yk) = pX(xj) pY(yk) for discrete X,Y
fX,Y(x,y) = fX(x) fY(y) for continuous X, Y
FX,Y(x,y) = P[X < x ,Y< y] = FX(x) FY(y)
If X and Y are independent, so are g(X), and
h(Y)
Example on marginal pmf. (discrete r.v.)
A random experiment consists of tossing 2 ‘loaded’ dice and noting the
pair of numbers (X, Y) facing up. The joint pmf pX,Y(j, k) is given as

k
j
1
2
3
4
5
6
1
2/42
1/42
1/42
1/42
1/42
1/42
2
1/42
2/42
1/42
1/42
1/42
1/42
3
1/42
1/42
2/42
1/42
1/42
1/42
4
1/42
1/42
1/42
2/42
1/42
1/42
5
1/42
1/42
1/42
1/42
2/42
1/42
6
1/42
1/42
1/42
1/42
1/42
2/42
6
pX(j) =  pX,Y(j,k) = 1/6
K=1
6
pY(k) =  pX,Y(j,k) = 1/6
j=1
for all ‘j’ 
for all ‘k’ 
6
 pX(j) = 1
j=1
6
 pY(k) = 1
K=1
Example of dependent r.v. (discrete)
In the earlier example,
pX(j) . pY(k) = 1/36
for all pairs (j, k)
But,
pX,Y(j, k) = 2/42 for j = k and
pX,Y(j, k) = 1/42 for j  k
pX,Y(j, k)  pX(j) . pY(k) for any pair (j, k)
 X and Y are NOT independent.
Example on marginal pdf. (continuous r.v.)
Find the normalization constant ‘c’ and the marginal pdf’s
for the joint pdf given below.

Ce-xe-y
0yx<
0
elsewhere
fX,Y(x,y) =
Normalization condition
 
1
  ce

 C=2
 x
x
e
y
dydx 
  ce
0 0
x
e
y
dydx 
c
2
Example continued …
The marginal pdf’s are given as

f X (x) 

x
f X ,Y ( x , y ) dy 


f X ,Y ( x , y ) dx 
x
 2e e
y
It can be verified that

0
dy  2 e
x
(1  e
x
)
0x<



y
0

fY ( y ) 
x
 2e e

f X ( x ) dx 

0
f Y ( y ) dy  1
y
dx  2 e
2 y
0y<
Example of dependent r.v. (continuous)
In the previous example
fX(x) . fY(y) = 4e-xe-2y(1-e-x)  2e-xe-y = fX,Y(x, y)
Thus the r.v.’s are NOT independent
Conditional Distribution and Density
Functions for Discrete r.v.
With Discrete X,Y given that Y = yk
Distribution Function:
FX|Y(x|yk) = P[Xx,Y=yk]
P[Y=yk]

Density Function
pX|Y(xj|yk) = pX,Y(xj|yk) = P[X = xj ,Y = yk]
Py(yk)
P[Y=yk]
Conditional Distribution and Density
Functions for Continuous r.v

With Continuous X and Y given fY(y)
Distribution Function:
x
FX|Y(x|y) = ∫ fX,Y(u,y)du
-∞
fY(y)
Density Function:
fX|Y(x|y) = fX,Y(x,y)
fY(y)
Example
Use of marginal pdf and joint pdf to get conditional pdf.
Let X and Y be the random variables with following joint pdf.
fX,Y(x,y) = 2 e-xe-y. ( 0  y  x <  and 0 otherwise. )
In an earlier example the marginal pdf ’s were found to be
fX(x) = 2 e-x(1 – e-x)
0  x <  and
fY(y) = 2 e-2y
0y<
Find their conditional pdf ’s.
Solution :f X (x | y) 
fY ( y | x ) 
f X ,Y ( x , y )
x

fY ( y )
f X ,Y ( x , y )
f X ( x)
2e e
2e
y
2 y
x
2e e

2e
x
e
y
(1  e
x
For x  y
( x y )

)
e
y
1 e
x
For 0 < y < x
Conditional Distributions & Independence
X and Y are independent if and only if the
conditional distribution function, the conditional
probability mass function, or the conditional
density function of X, given the Y = y, is identical
to the marginal function.
Conditional Expectation

Given that Y = y, the expectation of X is:
 Discrete Case:
∞
E[X|Y=y] = Σ xjpX|Y(xj|y)
j=1
Continuous∞Case:
E[X|Y=y] = ∫ x fX|Y(x|y)dx

-∞


The conditional expectation can be viewed as defining a
function of y : g(y) = E[X | y].
Hence, g(Y) = E[X | Y] is a random variable.
Properties of conditional expectation.

E[ E[X|Y] ] = E[X]
For the case of continuous r.v.s
Let g(Y) = E[X|Y], then


 

E  g (Y )    g ( y ) f Y ( y ) dy     xf X ( x | y ) dx  f Y ( y ) dy



   


 
  x   f X ( x | y ) f Y ( y ) dy

  


 xf
X
( x ) dx  E  X

 dx 



 
 x   f X ,Y ( x , y ) dy
  

dx




This is also true for any function of X. i.e.
E[ E[h(X)|Y] ] = E[ h(X) ]
 V[X] = E[X2] – (E[X])2 = E[E[X2|Y]]-(E[E[X|Y]])2.

Example.
The total number of defects X on a chip is a Poisson variable
with mean ‘’. Suppose that each defect has a probability ‘p’
of falling in a specific region ‘R’ and location of each defect is
independent of the location of any other defect. Find the pmf
of the number of defects Y that fall in the region ‘R’.

This is a case of discrete r.v.s. If ‘k’ is the total number of defects on
the chip and ‘j’ of them fall in the region ‘R’, the pmf for Y = j is given by

P [Y  j ] 


p X ,Y ( k , j ) 
k 0
Continued …
 P [Y
k 0
 j | X  k ]P[ X  k ]
---- eq. (I)
Example continued …
Now,
P[ Y=j | X=k ] = Probability that ‘j’ defects fall in region ‘R’, given that
totally there were ‘k’ defects on the chip.
This is a case of binomial distribution with parameters ‘k’ and ‘p’.
0,
j>k
P[ Y=j | X=k ] =
kC
j
0jk
pj (1 – p)K – j ,
Substituting, in eq(I) and noting that ‘X’ has Poisson distribution,

 P [Y
P [Y  j ] 
 j | X  k ]P[ X  k ]
k 0



k j
k
C j p (1  p )
j
k j

k
k!
e


( p )
j!
j
e
 p
Thus, the defects falling in
Region ‘R’ has a Poisson
distribution with Parameter
‘p’.
Example. (Conditional expectation)
In the last example, the number of defects falling in a
specific region ‘R’ (Y) was found to have Poisson distribution
with parameter ‘p’.
Hence, mean of ‘Y’ = p.
We can get the same result by using conditional expectation.


E [Y ]  E [ E [Y | X ]] 
 E [Y | X
 k ]P[ X  k ]
k 0


 ( kp ) P [ X
k 0

 k ]  p  kP [ X  k ]  pE [ X ]  p 
k 0
Conditional Variance

Definition,

V[X|Y] = E[(X-E[X|Y])2|Y]
Another form
V[X|Y] = E[X2|Y] – (E[X|Y])2
Using the above form,
E[ V[X|Y] ] = E[ E[X2|Y] ] – E[ (E[X|Y])2 ] ---- (i)
And the definition of variance,
V[ E[X|Y] ] = E[ (E[X|Y])2 ] – (E[ E[X|Y] ])2 ---- (ii)
Adding (i) and (ii) we get a useful result.
 E[V[X|Y]] + V[E[X|Y]] = E[E[X2|Y]] – (E[E[X|Y]])2 =
V[X]
Functions of random variables.
Let
Z = X/Y. Find the pdf of Z if X and Y are independent
and both exponentially distributed with mean one.
We can use conditional probability.
FZ(z|y) = P[Z  z |y] = P[X/y  z |y]
=
P[X  yz |y]
----- if y > 0
Using chain rule,
dF/dz = (dF/du) . (du/dz)
P[X  yz |y]
----- if y < 0
Here, u = yz
FX(yz |y)
----- if y > 0
1 - FX(yz |y)
----- if y < 0
=
yfX(yz | y)
----- if y > 0
fZ(z | y) =
= |y| fX(yz | y)
- yfX(yz | y)
----- if y < 0
Continued …
Example continued…
Now, the pdf of Z is given by,

fZ (z) 

| y |
f X ( yz | y ) f Y ( y ) dy 

| y |
f X ,Y ( yz , y ) dy

Using the fact that X and Y are independent and exponentially distributed
with mean one,

fZ (z) 

0
ye
 yz
e
y
dy 
1
1  z 
2
Z>0
Another example.
A system with standby redundancy has a single key
component in operation and a duplicate of that component in
standby mode. When the first component fails the second is put
into operation. Find the pdf of the lifetime of the standby
system if the components have independent exponentially
distributed lifetime with the same mean.

Let X and Y be the lifetimes of the two components. Then the system
lifetime ‘T’ is given by,
T=X+Y
The cdf of T is found by integrating the joint pdf of X and Y over the
region of plane corresponding to the event {T  t}.
 tx
FT ( t ) 

 
f X ,Y ( x , y ) dydx
Continued…
Example continued…
The pdf of T is obtained from differentiating the cdf. Further, X and Y are
independent. This gives,
f T (t ) 
d
dt

FT ( t ) 


f X ,Y ( x , t  x ) dx 


f X ( x ) f Y ( t  x ) dx

The two pdf’s in the integrand (exponentially distributed) are given as
e-x
x0
fX(x) =
0
x<0
fY(t - x) =
e-(t – x)
xt
0
x>t
t
These substitutions give
f T (t ) 
 e
0
 x
e
 (t  x )
dx   te
2
 t
Expected value of functions of r.v.s

E[X1 + X2 +…+Xn] = E[X1] + E[X2] +…+E[Xn]
Let X1,X2,…,Xn represent repeated measurements of the
same random quantity. Then these variables can be considered
iid (Independent Identically Distributed). This means, for i =
1,…,n
 All Xi are independent of each other.
 E[Xi] = E[X]
 V[Xi] = V[X]


E[X1 + X2 +…+Xn] = nE[X], for iid.

V[X1 + X2 +…+ Xn] = nV[X], for iid.
Joint moment and Covariance
 

 E[XjYk]
=
j
k
x y f X ,Y ( x , y ) dxdy
X, Y jointly continuous

x
i
j
i
k
y n p X ,Y ( x i , y n )
X and Y discrete
n
Above is the definition of jkth joint moment of X and Y
 If j = k = 1, it is known as correlation.
 If E[XY] = 0, they are said to be orthogonal.
 The jkth central moment of X and Y is
E[(X – E[X])j (Y – E[Y])k]
 In the definition of jkth central moment, j = 2 and k = 0
gives V[X] while j = 0 and k = 2 gives V[Y].
 In the definition of jkth central moment, j = 1 and k = 1
gives covariance of X and Y.
Properties of covariance and corr. coeff.
Covariance of independent variables is 0.
COV(X, Y) = E[(X – E[X]) (Y – E[Y])] = E[XY] – E[X]E[Y] = 0.
 If any of the random variables has mean 0 then
COV(X, Y) = E[XY].
 Covariance generalizes variance, but it can be negative.
E.g. if Y = - X, then COV(X, Y) = E[XY] – E[X]E[Y]
= E[- X2] + (E[X])2 = - V[X]  0.
 The correlation coefficient of X and Y is defined as,
X,Y = COV(X, Y)/XY,
where X and Y are STDs of X and Y respectively.
 We have
-1  X,Y  1
 Correlation coefficient of independent variables is 0, but the
converse is not true.

Example.
Let  be uniformly distributed in the interval (0,2). Define X
and Y as
X = cos
and
Y = sin.
Show that the correlation coefficient between X and Y is 0.

We have,
E[ X ] 
1
2
2
 cos  d 
0
0
Similarly it can be proved that E[Y] = 0. Now,
E [ XY ]  E [sin  cos  ] 
1
2
2
 sin  cos  d 
0
0
E[XY] – E[X]E[Y] = 0  COV(X, Y) = 0  X,Y = 0
But X and Y are not independent, since X2 + Y2 = 1
Sum of random number of r.v.s
N
SN 

Xk
k 1
Given above is the sum SN of Xi (i=1,…) iid; where N is chosen
randomly and independent of each Xi. For each i, E[Xi] = E[X]
and V[Xi] = V[X].
 Then, E[SN] = E[N]E[X]

From the properties of conditional expectation,
E[SN] = E[ E[SN | N] ] ------- Slide 23.
= E[ NE[X] ] ------- Xi are iid’s.
= E[N]E[X]
------- N is independent of each Xi.
This result is valid even if the Xi’s are not independent. They
only need to have same mean.

Continued…
Continued…

V[SN] = E[N]V[X] + V[N](E[X])2.
V[SN] = E[ V[SN | N] ] + V[ E[SN | N] ] ----- Slide 27.
Here, V[ E[SN | N] ] = V[NE[X]]
= E[N2E2[X]] – (E[NE[X]])2.
= E2[X] (E[N2] – (E[N])2)
= V[N] (E[X])2.
And E[ V[SN | N] ] = E[NV[X]]
----- Slide 32
= E[N]V[X]
Substituting in the first step above gives the expected result.
Mean Square Error(MSE)
Used when a r.v. X is estimated using another r.v. Y.
MSE = E[(X-g(Y))2]
 When g(Y) = a (constant), a = E[X] for min. error.
 When g(Y) = αY + β (a linear estimator)
α = E[XY] – E[X]E[Y] and β = E[X] - α E(Y) for min. error.
V(Y)
Another way of expressing linear estimator is,

 Y  E [Y ] 
ˆ
  E[ X ]
X   X ,Y  X 

Y



When g(Y) is non-linear, g(Y) = E[X|Y] for minimum error.
The best estimator is g(Y) = E[X|Y]. If X and Y both have
Gaussian distribution, the best estimator is equal to the linear
estimator.
Example
The amount of yearly rainfall in city 1 and city 2 is modeled
by a pair of jointly Gaussian r.v.s X and Y with the joint pdf
given by the equation below. Find the most likely value of X
given that we know Y = y (i.e. E[X |Y=y]).

f X ,Y ( x , y ) 

1

exp 
2
 2 (1   X ,Y )
 x  m
1
 
   1
2

 x  m1
  2  X ,Y 

 
1


2  1 2 1  
 y  m 2

 
2

  y  m2

  
2
 




2
X ,Y
Solution :- The marginal pdf of Y is found by integrating fX,Y over the
entire range of X. It is given by,
The marginal pdf of Y shows that it is
 ( y  m ) / 2
e
a Gaussian random variable with
fY ( y ) 
2  2
mean m2 and variance 22.
Continued …
2
2
2
2
2
 

 

Figures for the example.
Joint Gaussian pdf
Conditional pdf of X for a fixed value
Of y.
Solution continued …
Now, fX(x | y) = fX,Y(x,y) / fY(y). This can be shown to be,
f X (x | y) 

1
exp 
2
2
2
(
1


)


X ,Y
1
2

 
1
( y  m 2 )  m1  
 x   X ,Y

2

 
2  1 (1   X ,Y )
2
2
Hence, the conditional pdf of X is also Gaussian. It has a conditional
mean and conditional variance given by,
E[X |Y=y] = m1 + X,Y(1/2)(y – m2)
And
V[X |Y=y] = (1)2(1 – (X,Y)2)
------ (The answer to the question)
The conditional expectation found above, has an additional
interpretation; which is given in the next slide.
Interpretation.
Note that the conditional expectation found in the previous
solution, namely,
E[X |Y=y] = m1 + X,Y(1/2)(y – m2)
is a function of y.
 Replacing ‘y’ by ‘Y’ we generate a random variable, namely,
E[X | Y]. Also replacing m1 by E[X], m2 by E[Y], 1 by X,
and 2 by Y, we get the following result.
E[X | Y] = E[X] + X,Y(X/Y)(Y – E[Y])

We have thus proved with this example that the best
estimator (LHS) is equal to the linear estimator (RHS) for
jointly Gaussian r.v.s. (slide 38)

Sample mean
Let X be a random variable for which the mean, E[X] = , is
unknown. Let X1,…,Xn denote n independent repeated
measurements of X. Then X1,…,Xn are iid’s.
 The sample mean defined as follows is used to estimate E[X].

M
n

1
n

n
X
j
j 1
Mn itself is a random variable and E[Mn] = , since the Xi
are iid.
 If Sn = X1 + X2 + … + Xn, then Mn = Sn/n.
 V[Mn] = (1/n2)V[Sn] = V[X]/n
 V[Mn]  0 as n  
 Mn becomes a good estimator as n  

Continued …
Sample mean continued …
Using Chebyshev’s inequality,
P [| M n  E [ M n ] |  ] 
V [M n ]

2
Substituting for E[Mn], V[Mn] and taking the complement
probability we get,
P [| M n   |  ]  1 

n
2
2
----- (A)
This means for any choice of error  and probability (1 - ),
we can select the number of samples n so that Mn is within  of
the true mean with probability (1 - ) or greater.
 The quantity on the RHS of (A) gives a lower bound on
probability.

Example
A voltage of constant but unknown value is to be measured.
Each measurement Xj is a sum of the desired voltage ‘v’ and a
noise voltage Nj of 0 mean and STD of 1 microvolt. How many
measurements are required so that the probability that Mn is
within 1 microvolt of the true mean is at least 0.99?

We have Xj = v + Nj
With the assumption that Xj are iid for all j, we have
E[Xj] = v and V[Xj] = 1
We require  = 1
Substituting in (A) and replacing inequality with equality for lower bound
on probability, we get
0.99 = 1 – (V[Xj]/n2)
Solving this we get n = 100.
Weak law of large numbers.

Let X1, X2, … be a sequence of iid r.v.s with finite mean
E[X] = , then for  > 0,
lim P [| M n   |  ]  1
n 
The weak law of large numbers states that for a large enough
fixed value of n, the sample mean using n samples will be close
to the true mean with high probability.

Strong law of large numbers.
Let X1, X2, … be a sequence of iid r.v.s with finite mean 
and finite variance, then



P lim M n    1
n 
The strong law of large numbers states that with probability
1, every sequence of sample mean calculations will eventually
approach and stay close to E[X] = .
 The strong law of large numbers requires the variance to be
finite but the weak law of large numbers does not.

Central Limit Theorem.
Let Sn be the sum of n iid r.v.s with finite mean E[X] =  and
finite variance 2. Let Zn be the zero mean, unit variance r.v.
defined by
Zn = (Sn - n) /n, then

lim P [ Z n  z ] 
n 
1
2
z
e
2
x /2
dx

The summands Xj need to have finite mean and variance.
They can have any distribution.
 The resulting cdf of Zn approaches the cdf of a zero-mean,
unit variance Gaussian r.v.

Example
Suppose that orders at a restaurant are iid r.v.s with mean  =
$8 and STD  = $2. After how many orders can we be 90% sure
that the total spent by all customers is more than $1000?

Let Xk denote the expenditure of the kth customer. Then the total spent by
‘n’ customers is,
Sn = X1 + X2 + … + Xn.
We have,
E[Sn] = 8n
and
V[Sn] = 4n
The problem is to find the minimum value of ‘n’ for which
P[Sn > 1000] = 0.90
With Zn as defined in the previous slide,
P[Sn > 1000] = P[Zn > (1000 – 8n)/2n] = 0.90
Continued …
Solution continued…
Since Zn is a Gaussian r.v. with mean 0 and variance 1, its pdf is given by
fZn (z) 
1
2
2
e
z /2
The given probability is then expressed as the following integral.
0 . 90  P [ Z n  z ] 
1
2

e
2
z /2
dz
z
Where, z = (1000 – 8n)/2n
The value of ‘z’ (- 1.2815) is found from the table 3.4 in Leon –Garcia
and the minimum value of ‘n’ is found by solving the quadratic equation
in n, namely,
8n – 1.2815(2)n – 1000 = 0.
The positive root of this quadratic equation gives n = 128.6
Thus after minimum 129 orders we can be 90% sure that the total spent
by customers is more than $1000.
Questions???