Applied Statistics | Assignment

APPLIED STATISTICS
Problem 1
Part i
Let g.f ∈R3 such that the vector product of g * f in parts is read as;
g*f =
[ g 2 f 3−¿ g 3 f 2
g3 f 1−¿ g 1 f 3
g 1 f 2−¿ g 2 f 1 ]
We can see from the above equality that :
[gx]=
[ 0 −g 3 g 2
g 3 0 −g 1
−g 2 g 1 0 ]
This satisfies the condition ;
g*f=[gx]f
Hence g is mapped such that g← →[gx]
From the mapping ,it confirms that g is linear and invertible.
Thus g=-g i.e a skew-symmetric
The elements in the main diagonals in the matrix are zero.
The product of I and the matrix is invertible.
The standard inner product on Rn is given as ;
(g(x),f)=-(f,g) ,
Hence (g(x),f)=-(f(x),g),for all x,f∈ R∞
Part ii)
From the scalar products properties,For all values of g ∈Rn
||f.g||2=f*g.f*g =g.fT*f.g=||g||2
Taking f as a ratio matrix,consider fT.f=1

You realize that in f we assume that ʎ are Eigen values of f, then g must be
the corresponding vectors:
||f.g||=|ʎg||=|ʎ|||g||
But ʎ=1 ,hence ʎ lies on the unit circle of a complex plane.
This impl,1 ies that ;
f*g=ʎg→fg- = fg= ʎw =ʎ w
It is noted tha f has got eigenvalues.
Eig(f)={eix,e-ic,1} or eig(f)={e-ic,e-ic,1}
Cconsider c∈ [ 0,1 ] , eig (f )={eic,e-ic,-1} and
We get det(f)=-1
Such that
Eig(f)={eic,e-ic,1} for c ∈[0,1]
Thus ʎ =1
There are two in Eigen values i.e ʎ=1 or 3
Problem 2
Part i
L(f(x))=((1-x2)f’(x)
∫
0
x
¿
¿ ¿)(f’(t)
)2 +k(f(t)tdt=(1-t2)f’( t)f(t))|1-1 +∫
−1
1
¿ ¿
)|1-1
limx→1-(1-x2)f’(x)f(x)=∞
For some value greater than 0 and stretches near -1,we have
(1+x2f’(x)f(x)>k
You can assume that (1-x2)f’(x)>0 and f(x)<0 and near -1,hence Ƹ
∈
(0,1)

|((1-x2)f’(x))|f(x)>k ¿( ( 1−x2 ) f ' ( x ))∨¿
¿ ¿
(x∈ {, 1¿ ¿
You are to consider so as to you integrate for all x ∈[Ƹ,1]
∫
Ƹ
∞
¿ ¿ ¿
≥
k|
¿((1−t2 ¿ f ' 9 t ))∨dt ¿
( 1−t2 ) f ' ( t )
¿ ¿
dt
=k|In(1-t2)f’(t))|∞
Ƹ
≥
k|In(1-x2)f’(x)|
You realize that for some particular asset and property k for the
parameter to (0,k).The right-hand-side of the result tend to ∞ as
x→1-.Similarly,because f and [f] belong to L2(-1,1],convergence
occurs on the left-hand side as x →1-
Hence limx→1-(1-x2)f’(x)g(x)=L
When L=0,we suppose(1-x2)f’(x)>0 near 1- and repeat the
argument for the some k>0
Then,
∫
Ƹ
x
¿ ¿
¿ ¿|dt≥k|In((1-x2)f’(x)|x∈(Ƹ,1)
Part ii
Find the orthonormal basis Ƹ that constitutes e1,e2,,.....en whereby
e1,e2,,.....en are eigen values of L.
Assume that when n=1,V is one dimensional.
Let V be a vector space and L the operator defined by m+1 such
that dim V =n=m+1