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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 .

- Non - Differentiability of at .
- is non differentiable on set and is non differentiable on set , then is non differentiable on .
- Minimization of at . In other words, the
**Median minimizes the Sum of Absolute Deviation**.

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 .

- Non - Differentiability of at .
- is non differentiable on set and is non differentiable on set , then is non differentiable on .
- Minimization of at . In other words, the
**Median minimizes the Sum of Absolute Deviation**.

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!

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