While heading to a conference on board a train, three statisticians meet three biologists. The biologists complain about the cost of the train fare, but the statisticians reveal a cost-saving trick. As soon as they hear the inspector’s voice, the statisticians squeeze into the toilet. The inspector knocks on the toilet door and shouts: “Tickets, please!” The statisticians pass a single ticket under the door, and the inspector stamps it and returns it. The biologists are impressed. Two days later, on the return train, the biologists showed the statisticians that they have bought only one ticket, but the statisticians reply: “Well, we have no ticket at all.” Before they can ask any questions, the inspector’s voice is heard in the distance. This time the biologists bundle into the toilet. One of the statisticians secretly follows them, knocks on the toilet door and asks: “Tickets please!” The biologists slip the ticket under the door. The statistician takes the ticket, dashes into a another toilet with his colleagues, and waits for the real inspector. The moral of the story is simple: “Don’t use a statistical technique that you don’t understand.”

Filed under: Random Tagged: Jokes, Statistics ]]>

Let . Prove that as ,

a) the distribution of converges to for some real number ,

b) but converges to 2.

(*Credits:* I learnt of this problem from Persi Diaconis in my probability class.)

For each , let be the * simple truncation* of , i.e. , or equivalently

It is easy to calculate the following:

If we let , then

Let us check that * Lyapounov’s condition* holds for :

Since , as , the denominator of the above goes to infinity, which means that the entire RHS goes to zero, i.e. Lyapounov’s condition holds. Hence, we can use * Lyapounov’s Central Limit Theorem* to conclude that

(Here, means “converges in distribution”.) Next, note that

By the * first Borel-Cantelli lemma*, we conclude that . From this, we can further conclude that

where .

It is straightforward to show that converges to 2:

**Done!**

Filed under: Undergraduate Tagged: Probability ]]>

For , define the random variable

Let . Prove that as ,

a) the distribution of converges to for some real number ,

b) but converges to 2.

**Hint:** Simple truncation. Let . Use Lyapounov’s Central Limit Theorem and Borel-Cantelli’s 1st Lemma.

Filed under: Undergraduate Tagged: Probability ]]>

Let . Prove that as ,

a) the distribution of converges to for some real number ,

b) but converges to 2.

(*Credits:* I learnt of this problem from Persi Diaconis in my probability class.)

Filed under: Undergraduate Tagged: Probability ]]>

Show that the infinite series converges and find its sum.

The recurrence relation is begging to be * exponentiated* so that we have on both sides:

and now this looks like a * telescoping series* that allows cancellation of terms:

To show that converges, we have to show that exists. There are many different techniques to show that a limit exists, we can decide which is appropriate by playing around with the sequence.

After working out the first few terms of the sequence, it may become clear that * the sequence is decreasing*. The recurrence relation seems to suggest it as well: if we can state with confidence that is non-negative, then

It seems reasonable that should be positive as well. Let’s try using * induction to prove both statements at once*:

**Induction statement** : For , and .

It is clear that is true. Let us assume that is true for some . To prove , we can work backwards:

which is a true statement for all real numbers ! Hence the sequence is non-negative and we can use the earlier argument to show that . We have shown that , so by induction, the induction statement is true.

* Since we have a decreasing sequence with a fixed lower bound, it means that the sequence converges.* It remains to figure out what the limit is. We can use the recurrence relation to figure that out:

Hence, , and

**Done!**

Filed under: Undergraduate, USA Tagged: 2016, Algebra, Putnam, Sequences ]]>

Show that the infinite series converges and find its sum.

Filed under: Undergraduate, USA Tagged: 2016, Algebra, Putnam, Sequences ]]>

In the meantime, I thought I’d share the data I’ve scraped so far so that others who have the time and interest might have a go at analysing the data. Data is available at my Github repo.

All data was scraped from imo-official.org; the scripts I used to scrape them are in the ETL folder of the same repo. The data generally looks clean except some minor issues for “Contestant” (i.e. names of contestants). For example, my name is written in one order for 2003 and 2005 but in a different way for 2004. I have no idea how widespread this issue is, although a cursory glance at contestants in my country suggest the issue is a minor one.

Filed under: Grade 12, Intl/Regional Tagged: IMO ]]>

* Congratulations to my country, Singapore’s team, for coming in 4th overall!* And for winning the country’s:

**Best**medal haul (4G 2S, previous best was 4G 1S 1B in 2011)**Highest**absolute score (196 out of a possible 252, previous best was 179 in 2011)**2nd best**overall team ranking (4th, best was 3rd in 2011)

Some factoids on the results:

- The Gold, Silver and Bronze medal score cutoffs were
**29, 22 and 16**respectively. Nothing too out of the ordinary. (Interestingly, IMO 2014 had the exact same cutoffs.) - The top 3 countries were the
**United States, South Korea and China**. (These 3 countries were the top 3 in 2015 as well, just a different order: USA, China, South Korea.)- Team USA won a best possible 6 Gold medals. (Next best was 4G 2S by South Korea, China and Singapore.)

- There were
**6**contestants with perfect scores: 3 from South Korea, 2 from USA and 1 from China.- 3 of them were first-time participants of the competition. 2 of them had won Gold in 2015, while the remaining contestant (Allen Liu, USA) had won Gold in 2014 and 2015.

**Question 3**was the most difficult question, with a mean score of 0.251, 91% of students scoring 0 points and only 10 contestants scoring a perfect 7 points. The easiest was**Question 1**with a mean score of 5.272. (The questions from easiest to hardest: 1, 4, 2, 5, 6, 3.)

And finally, some useful links:

- IMO 2016 official website (At the time of writing of this post, the results have not been posted on this website.)
- AoPS forum on Contests and Problem Sets (All the latest threads are related to IMO 2016.)
- imo-official.org page for IMO 2016 (Contains all the stats you could dream of for IMO 2016.)

Filed under: Grade 12, Intl/Regional Tagged: 2016, IMO ]]>

(a) ,

(b) there exist integers , , and , and prime where ,

(c) divides , , and , and

(d) each ordered triple and each ordered triple form arithmetic sequences.

Find .

* There are many variables in this problem, with some of them ordered in a non-traditional way (see (b)).* Let’s try to lay them out in a way that makes sense:

Here, we’ve put (a) and (b) in a diagram. * The next bit of information we’ll put into the diagram is (d).* Since we are working with triplets forming arithmetic sequences, I prefer to have them in the form instead of . This is because the former has symmetry which allows it to sum to , while the latter sums to the uglier . Let’s put this information in:

(Note that and must both be positive, and that .) Now, let’s write out the 3 expressions that are supposed to be divisible by :

* A natural next move is to manipulate the relations , and to isolate variables on the right; this will (hopefully) restrict the choices for . *There are a lot of similarities on the RHSes of the 3 relations so we can hope that something simple comes out of it.

gives

and gives

The term is repeated in the 2 relations above! We can subtract one from the other to obtain

Remember that ? This means that is relatively prime to , and so

, i.e.

* This is a big result!* Because of condition (b), we must have , , and . Let’s update our variable diagram:

The new is the easiest to make sense of: must be a multiple of 3. For each value of , we can list the admissible values of and :

At first, as increases, the number of possible triplets increases. For each , the thing that is stopping more possible triplets is hitting up against the constraint that . However, at some point as gets too big, the constraint we hit up against first is .

* Where does this happen?* One might imagine that it happens somewhere in the middle:

If , then when , we have which is **admissible**. Hence, the constraint we hit first is still .

If , then when , we have which is **inadmissible**. Hence the constraint we hit first is now .

Now, note that every time increases from 3 to 48, the number of possibilities increases in increments of 1, from 1 to 16. In the other direction, as decreases from 96 to 51, the number of possibilities increases in increments of 1, from 1 to 16. Putting all this together, the number of possible triplets is

**Done!**

Filed under: Grade 12, USA Tagged: 2013, AIME, Number Theory ]]>

(a) ,

(b) there exist integers , , and , and prime where ,

(c) divides , , and , and

(d) each ordered triple and each ordered triple form arithmetic sequences.

Find .

Filed under: Grade 12, USA Tagged: 2013, AIME, Number Theory ]]>