This post has a two-faced problem, which gives you both an analytical and statistical insight into ISI MStat 2013 PSB Problem 1.
If and also
, where
is any real number then prove that
for all
and
.
has non - differentiability at
, since each of
is non differentiable at
. The result follows from the first two results in the pre-requisites.
Now, since, , the set of non differentiability points of
is
while the set of non differentiability points of
is
.
.
Thus, {} = {
}.
Therfore, for all
and
.
The Median Minimizes the Sum of Absolute Deviation has really elegant proof here.
We will see the idea from a different perspective with respect to the diagrams and use the idea to prove the actual problem.
For the set { 1 < 2 < 3 < 4 < 5 }, the median is 3. Hence, the is minized at
.
But for the set { 1 < 2 < 3 < 4 < 5 < 6}, the median is . Rather the
is minized at
. Let's for the time being call
as the median of { 1 < 2 < 3 < 4 < 5 < 6}.
Now, the next idea is simple yet elegant.
= {
};
= {
}.
=
;
=
.
=
.
Step 1 : Take the minimum of and
. They must be equal.
Step 2 : Therefore, median of = median o f
.
Step 3 : median of
;
median of
. This means, we delete the equal elements in both
and
.
Step 4 : Go to Step 1. Stop if
This will converge only if .
Hence, for all
and
.
Thus, we gave two different views of the same problem.
Hope you enjoyed it.
Stay Tuned!
This post has a two-faced problem, which gives you both an analytical and statistical insight into ISI MStat 2013 PSB Problem 1.
If and also
, where
is any real number then prove that
for all
and
.
has non - differentiability at
, since each of
is non differentiable at
. The result follows from the first two results in the pre-requisites.
Now, since, , the set of non differentiability points of
is
while the set of non differentiability points of
is
.
.
Thus, {} = {
}.
Therfore, for all
and
.
The Median Minimizes the Sum of Absolute Deviation has really elegant proof here.
We will see the idea from a different perspective with respect to the diagrams and use the idea to prove the actual problem.
For the set { 1 < 2 < 3 < 4 < 5 }, the median is 3. Hence, the is minized at
.
But for the set { 1 < 2 < 3 < 4 < 5 < 6}, the median is . Rather the
is minized at
. Let's for the time being call
as the median of { 1 < 2 < 3 < 4 < 5 < 6}.
Now, the next idea is simple yet elegant.
= {
};
= {
}.
=
;
=
.
=
.
Step 1 : Take the minimum of and
. They must be equal.
Step 2 : Therefore, median of = median o f
.
Step 3 : median of
;
median of
. This means, we delete the equal elements in both
and
.
Step 4 : Go to Step 1. Stop if
This will converge only if .
Hence, for all
and
.
Thus, we gave two different views of the same problem.
Hope you enjoyed it.
Stay Tuned!