June | 2012 | Chaitanya's Random Pages

June 26, 2012

Exports and Imports by Country

Filed under: geography — ckrao @ 1:08 pm

I recently had a look at the Year Book Australia, 2012 and was interested in seeing the current state of Australian Industry. From this page on the Australian Bureau of Statistics website, in 2010-2011 Australia had a GDP of around 1.32 trillion Australian dollars ($58k per capita) with the following breakdown by sector.

One can see here how the Construction and Financial and Insurance sectors have increased in GDP share over the past decade while manufacturing has decreased. Interestingly Mining is only around 7% of GDP and this figure has remained steady from ten years ago.

In mining as of 2009 Australia was the world’s largest “producer” of iron ore, and fourth largest (behind China, USA, India) of coal. Interestingly four countries produce 75% of the world’s iron ore: Australia, Brazil, India and China.

In 2009-2010 Australia had a business expenditure on Research and Development of around 1.3% of its GDP on (source: here)

Imports and exports were each of value around 20% of GDP (source: here). I stumbled onto the Observatory of Economic Complexity that helps one to visualise the breakdown further, over the last 15 years.

Compare the 1995 and 2010 breakdowns to see how China has become Australia’s largest export partner.

Countries Australia exports to in 1995 (left) and 2010 (right)

Next we see a breakdown of Australian exports and imports by sector. Note how dominant the mineral products have become.

Australian Exports in 1995 (left) and 2010 (right)

Finally we see machinery and transportation dominating Australian imports.

Australian Imports in 1995 (left) and 2010 (right)

For more such tree maps, see this list of countries by economic complexity.

June 23, 2012

The shortest day of the year

Filed under: climate and weather,mathematics — ckrao @ 1:14 pm

June 21 was the shortest day of this year here in Melbourne, Australia and I saw on timeanddate.com that the length of the day was 9h 32m 31s. I was curious to see how close this was to the rough estimate that one can derive for the length of day on the winter solstice based on the calculation I give below. We make the following simplifying assumptions.

The earth is spherical
On the winter solstice for the southern hemisphere the sun is directly over the tropic of cancer (i.e. directly overhead there at some time during the day)

We use the figure below to visualise the situation at this time. Assume the sun is a long way to the right of screen. It shows the sun directly overhead at the tropic of cancer (C) and Melbourne at solar noon. The left half of the image experiences night. Imagine the earth rotating about the axis joining the north and south poles (NS). Melbourne will then rotate around this axis and in this image it will always be somewhere on the line passing through D and M. When it is at the point D it corresponds to dawn or dusk. Hence the length of its daylight will the time in which it is in the red region.

Note that the angle $\theta$ is equal to the earth’s angle of tilt, which is known to be approximately 23.5 degrees.

Below is another look at the red region from a different angle of perspective. In this figure below the line $DQD'$ corresponds to the single point $D$ in the figure above. Let $\alpha = \angle MPD$ . Then the fraction of the day in daylight will be $\alpha/\pi$ where $\alpha$ is measured in radians.

Using the notation below, the daylight time will be $\alpha/\pi \times 24$ hours.

We use trigonometry to find $\alpha$ . Firstly in triangle $OPM$ ,

$\displaystyle r = R \cos \phi \quad \quad (1),$

where $\phi$ represents the latitude of Melbourne (38 degrees).

Secondly $OP = R \sin \phi$ . Hence

$\displaystyle PQ = OP \tan \theta = R \sin \phi \tan \theta. \quad \quad (2)$

Finally, in triangle $PDQ$ , $\cos \alpha = PQ/r$ so combining this with (1) and (2) gives

$\displaystyle \cos \alpha = \frac{R \sin \phi \tan \theta}{ R \cos \phi} = \frac{\sin \phi \tan \theta}{\cos \phi}.$

Hence $\alpha = \arccos\left(\tan \phi \tan \theta\right)$ and for Melbourne the predicted length of daylight is

$\displaystyle \alpha/\pi \times 24 = \frac{24 \arccos\left(\tan 37.783^{\circ} \tan 23.438^{\circ}\right)}{\pi} \approx$ 9h 22m 54s.

This is about 10 minutes (1.8%) shorter than the actual time, the difference being due to effects such as the sun not being a point object, diffraction of light through the atmosphere and the approximations listed above. However the above calculation is a handy guide. A more precise calculation is shown here.

Incidentally, the shortest day here this year was curious temperature-wise as the graph below shows (data from the Australian Bureau of Meteorology).

The temperature hovered between 9.5°C and 9.9°C (at the half-hour sampling instances) between 6:30am and 7:30pm, with higher temperatures on either side! The maximum (after 9am) of 10.5°C was only reached at around 10pm.

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June 11, 2012

Curious high or low temperature spots

Filed under: climate and weather,geography — ckrao @ 12:17 am

Here I provide a list of some places that go against our intuition about being surprisingly hot or cold at a certain time of the year given their latitude. Most of the places chosen are close enough to sea level so that elevation is not the cause (more likely ocean or continental influence).

The east coast of Australia up to Sydney and Newcastle has a mid summer average maximum temperature of only 25.5°C – highly unusual for the east coast of a continent. At a corresponding latitude (33° from equator) the mid summer maximum at the east coast of Asia (Shanghai at 31°N) is 31.8°C, of South America (Buenos Aires at 34.6°S) is 30.4°C and in North America (Charleston, NC, USA) is 32.8°C.
Walvis Bay, Namibia is in the tropics (23°S) yet has a mid summer average maximum temperature of only 22°C! Western South America is similarly cool with a mid summer average maximum of 25°C (Antofagasta, Chile). The corresponding west coast place in Australia has an average maximum of 32.4°C (Carnarvon at 25°S), in Africa 27°C (Dakhla, Western Sahara at 23.7°N), and in North America 32.4°C (Mazatlán, Mexico).
Eureka, CA (USA) at 40.8°N has an average maximum of only 17.7°C in its warmest month. Contrast this with Fairbanks, Alaska at 65°N which still reaches an average 23°C in its warmest month with an all-time high of 37°C!
Kyzyl, the capital of the Tuva republic in Russia, has an average maximum of 27°C in July and average minimum of -35°C in January! London is at the same latitude.
Iceland and Ireland have very low variation between summer and winter temperatures due to ocean currents of the East Atlantic. Reykjavik, Iceland averages 2°C maximum in its coldest month up to 13°C in its warmest.
The Turpan depression in northwest China surely has the hottest summer temperatures for a place so far from the equator. For example Turpan at 43°N has an average maximum of 40°C in July! A similar hot spot is Ashgabat, Turkmenistan (38°N) with an average maximum of 38°C in July. It has the same distance from the equator as Melbourne, Australia which only has an average maximum of 26°C in mid summer.
Tromsø, Norway is at 69.7°N (hence beyond the arctic circle) yet in its coldest month its average temperature range is -6.5 to -2.2°C.
Some parts of southern China are surprisingly cool for their latitude in winter. For example Guilin in Guangxi province is at 25°N and only 150m in elevation, yet has a range of 5.4 to 11.5°C in January (cooler than Tromsø’s warmest month!)
Lima, Peru is coastal, only 12° south of the equator, yet has a maximum temperature range from 18.4° to 26.5°C from coolest to warmest months. Contrast this with Chennai, India which is also coastal and whose corresponding range is 29° to 38°C.
The Southern Ocean shows very little variation between summer and winter temperatures. For example Bird Island in South Georgia at 54°S ranges from an average minimum of -5.4°C in its coldest month to an average maximum of 5.6°C in its warmest month.

Below is a map of showing all the places mentioned in this post.

June 10, 2012

Outline proof of the extreme value theorem in statistics

Filed under: mathematics — ckrao @ 9:15 am

Recently I read a proof in [1] of the main theorem of extreme value statistics: the Fisher–Tippett–Gnedenko theorem. In this post I give an outline.

Here we are interested in the maximum of many independent and identically distributed random variables $X_i$ with distribution function $F$ . Let $x^* = \sup \{x: F(x) < 1\}$ which may be infinite. Then as $n \rightarrow \infty$ ,

${\rm Pr}(\max(X_1, X_2, \ldots X_n) \leq x) = F^n(x)$ tends to 0 if $x < x^*$ and 1 if $x > x^*$ .

Therefore $\displaystyle \max(X_1, X_2, \ldots, X_n)$ converges in probability to $x^*$ as $n \rightarrow \infty$ .

In order to avoid this degenerate limiting distribution for all extreme value distributions, it is necessary to normalise the distribution.

To this end, suppose there exist real numbers $a_n >0, b_n$ such that

$\displaystyle \frac{\max(X_1, X_2, \ldots, X_n) - b_n}{a_n}$ approaches a non-degenerate limiting distribution.

In other words, there exists a distribution function $G(x)$ such that

$\displaystyle \lim_{n \rightarrow \infty} F^n(a_n x + b_n) = G(x).$

Taking logarithms of both sides, this is equivalent to

$\displaystyle \lim_{n \rightarrow \infty} n \log F(a_n x + b_n) = \log G(x)$

This requires $F(a_n x + b_n) \rightarrow 1$ as $n \rightarrow \infty$ . Using $\log x \approx x -1$ for $x$ close to 1, the above is also equivalent to

$\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n(1 - F(a_n x + b_n) } = \frac{1}{- \log G(x)}. \quad \quad (1)$

Next we use the following definition.

A non-decreasing function $f$ has left-continuous inverse $f^{\leftarrow}$ defined by

$\displaystyle f^{\leftarrow}(x) := \inf\{y: f(y) > x\}.$

One can use this definition to prove

Lemma 1: If $f_n(x) \rightarrow g(x)$ for non-decreasing functions $f_n$ then for each continuity point $x$ we have $f_n^{\leftarrow}(x) \rightarrow g^{\leftarrow}(x)$ .

Next we claim that (1) is equivalent to

$\displaystyle \frac{U(nx) - b_n}{a_n} \rightarrow G^{\leftarrow}\left(e^{-1/x}\right) \quad \quad (2)$

where $U$ is the left-continuous inverse of $\frac{1}{1-F}$ . To see this, let $V(x) = \frac{1}{1 - F(x)}$ . Then by the definition of $U$ , $U(x) = \inf \{ y: V(y) \geq x \}$ . Then for any $n \in \mathbb{N}$ ,

$U(nx) = \inf \left\{ y: \frac{1}{n(1-F(y))} \geq x \right\}$

and so

$\begin{array}{lcl} \frac{U(nx) - b_n}{a_n} &=& \inf \left\{\frac{y - b_n}{a_n}: \frac{1}{n(1 - F(y))} \geq x \right\}\\&=& \inf \left\{z: \frac{1}{n(1 - F(a_n z + b_n))} \geq x \right\}\end{array}$

By (1) and the lemma as $n \rightarrow \infty$ this tends to

$\begin{array}{lcl} & & \inf \left\{ z: \frac{1}{-\log G(z)} \geq x \right\}\\& = & \inf \left\{ z: \log G(z) \geq \frac{-1}{x} \right\} \\ &=& \inf \left\{ z: G(z) \geq e^{-1/x} \right\}\\ &=& G^{\leftarrow} \left( e^{-1/x} \right), \end{array}$

proving the claim. We can also write

$\displaystyle \lim_{t \rightarrow \infty} \frac{U(tx) - b(t)}{a(t)} = G^{\leftarrow}\left(e^{-1/x}\right) =: D(x) \quad \quad (3)$

where $a(t) := a_{[t]}, b(t) := b_{[t]}$ and $[t]$ denotes the integer part of $t$ .

We are now ready to prove the main theorem of extreme value theory.

Theorem (Fisher, Tippet, Gnedenko):

$\displaystyle G_{\gamma}(ax + b) = \exp\left( -(1 + \gamma x)^{-1/\gamma} \right), \quad 1 + \gamma x > 0, \quad \gamma \in \mathbb{R}$

where the right side is equal to its limiting value $\exp \left( - e^{-x}\right)$ if $\gamma = 0$ .

Proof:

This will involve numerous substitutions but the main idea is to arrive at a differential equation that can be solved to obtain the above. Suppose $1$ is a continuity point of $D$ . Then for any continuity point $x > 0$ ,

$\displaystyle \lim_{t \rightarrow \infty} \frac{U(tx) - U(t)}{a(t)} = D(x) - D(1) =: E(x). \quad \quad (4)$

We can write

$\displaystyle \frac{ U(txy) - U(t)}{a(t)} = \frac{U(txy) - U(ty)}{a(ty)} \frac{a(ty)}{a(t)} + \frac{U(ty) - U(t)}{a(t)}.$

The claim is that both $\frac{a(ty)}{a(t)}$ and $\frac{U(ty) - U(t)}{a(t)}$ have limits as $t \rightarrow \infty$ . If they had more than one limit point, say $A_1, A_2, B_2, B_2$ then for $i = 1, 2$ (4) in the limit $t \rightarrow \infty$ gives us

$E(xy) = E(x) A_i + B_i$ .

Subtracting gives $E(x) (A_1 - A_2) = B_2 - B_1$ which implies $A_1 = A_2, B_2 = B_1$ as we know $E(x)$ is non-constant (since we seek a non-degenerate solution).

We conclude that

$\displaystyle E(xy) = E(x) A(y) + E(y). \quad \quad (5)$

This is a functional equation that we wish to solve. We let $s:= \log x, t := \log y, H(x) := E(e^x)$ to obtain

$\displaystyle H(t+ s) = H(s) A(e^t) + H(t),$

which using $H(0) = E(1) = 0$ implies

$\displaystyle \frac{H(t+s) - H(t)}{s} = \frac{H(s) - H(0)}{s} A(e^t). \quad \quad (6)$

Since $H$ is monotone (following from the monotonicity of $D$ ), it is differentiable at some $t$ . By (6) it is differentiable at all $t$ . Indeed from (6) we obtain

$\displaystyle H'(t) = H'(0) A(e^t) \neq 0. \quad \quad (7)$

Let $Q(t) = H(t)/H'(0)$ . Then $Q(0) = 0, Q'(0) = 1$ .

From (6) and (7),

$\displaystyle Q(t+s) - Q(t) = Q(s) A(e^t) = Q(s) H'(t)/H'(0) = Q(s) Q'(t). \quad \quad (8)$

Similarly, $Q(s+t) - Q(s) = Q(t) Q'(s)$ and upon subtraction from (8) we obtain $Q(s) - Q(s)Q'(t) = Q(t) - Q(t)Q'(s)$ from which

$\displaystyle \frac{Q(s)}{s} (Q'(t) - 1) = Q(t) \frac{Q'(s) - 1}{s}.$

Taking the limit as $s \rightarrow 0$ and using $Q'(0) = H'(0)/H'(0) = 1$ gives the following differential equation for $Q$ .

$\displaystyle Q'(t) -1 = Q(t) Q''(0), Q(0) = 0, Q'(0) = 1 \quad \quad (9)$

To solve (9), differentiate both sides with respect to $t$ : from $Q''(t) = Q'(t) Q''(0)$ we see that

$\displaystyle (\log Q')' (t) = Q''(0) =: \gamma$ .

Hence $Q'(t) = e^{\gamma t}$ and since $Q(0) = 0$ , $Q(t) = \int_0^t e^{\gamma s}\ ds$ .

Recalling $Q(t) = H(t)/H'(0)$ , this leads to $H(t) = H'(0) \frac{e^{\gamma t} - 1}{\gamma}$ .

Recalling $H(x) := E(e^x) := D(e^x) - D(1)$ this means

$D(t) = D(1) + H'(0) \frac{t^{\gamma} - 1}{\gamma}.$

Taking the left-continuous inverse of both sides,

$D^{\leftarrow}(x) = \left(1 + \gamma \frac{x - D(1)}{H'(0)} \right)^{1/\gamma}.$

Since $D(x) := G^{\leftarrow}\left(e^{-1/x}\right)$ , $D^{\leftarrow}(x) = \frac{-1}{\log G(x)}$ .

Hence

$\displaystyle G(x) = e^{-1/D^{\leftarrow}(x) }.$

In other words,

$\displaystyle G(H'(0) y + D(1) ) = \exp \left(- (1 + \gamma y)^{-1/\gamma} \right),$

where $\gamma = Q''(0) = H''(0)/H(0)$ .

If 1 is not a continuity point, we repeat the above proof with $U(t)$ replaced by $U(tx_0)$ where $x_0$ is a continuity point. This completes the proof.

This generalised extreme value distribution encaptures three distributions depending on the nature of the tail of the original distribution $X_i$ :

Type I – $\gamma = 0$ : Gumbel (double exponential) distribution (exponential tail – e.g. normal or exponential distribution)
Type II – $\gamma > 0$ : Fréchet distribution (polynomial tail – e.g. power law distribution to model extreme flood levels or high incomes)
Type III – $\gamma < 0$ : reverse-Weibull distribution (finite maximum – e.g. uniform distribution)

Sample density functions are plotted below for specific values of $\gamma$ .

Reference

[1] L. De Haan, A. Ferreira, Extreme Value Theory: An Introduction, Springer, 2006.

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June 7, 2012

Chris Gayle in recent T20s

Filed under: cricket,sport — ckrao @ 12:26 pm

Chris Gayle has amassed an unbelievable record in Twenty20 cricket. In 105 innings he has scored 3970 runs at an average of 43.62 with 8 hundreds and 26 fifties. Such statistics would be respectable in test cricket and fantastic for one-day internationals. Now consider that he has done this with a strike rate of 155.93 when at most 20 overs per innings are available! His statistics in the last 7 tournaments are even more mind boggling. It’s only a shame he hasn’t played a T20 international since 2010.

ESPN Cricinfo links:

The Best Batsman in Twenty20 Cricket

Chris Gayle in Twenty20 matches since 2011:A phenomenon in the shortest form

Below are his series-by-series scores since Jan 2011 together with some points of note.

1) 2010-11 Big Bash in Australia (played for Western Australia)
5th highest run scorer: 5 innings, 182 runs, HS 92, Ave 36.4, SR 193.61, 2 50s, 18 4s, 13 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Dec 30 2010	Tasmania	22	19	3	0	115.78
Jan 9 2011	NSW	61	30	8	4	203.33	opening partnership of 101 with Marsh in 8.3 overs 7th over: 6 6 6 4 4 6
Jan 13 2011	SA	6	2	0	1	300
Jan 18 2011	Vic	1	3	0	0	33.33
Jan 25 2011	Qld	92	40	7	8	230	opening stand 144 in 12.5 overs 120m six hit 50 reached off 20 balls

2) IPL 2011 in India (played for Royal Challengers Bangalore)
highest run scorer: 12 innings, 608 runs, Ave 67.55, SR 183.13, 2 100s, 3 50s, 56 4s, 44 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Apr 22 2011	Kochi Tuskers Kerala	102*	55	10	7	185.45
Apr 26 2011	Delhi Daredevils	26	14	4	1	185.71
Apr 29 2011	Pune Warriors	49	26	4	4	188.46
May 6 2011	Kings XI Punjab	107	49	10	9	218.36	he was 6 off 13 after 3 overs, then made 76 off the next 24 century off 46 balls
May 8 2011	Kochi Tuskers Kerala	44	16	3	5	275	opening stand 67 in 3.5 overs, 3rd over for 37 runs went 6, (nb)6, 4, 4, 6, 6, 4
May 11 2011	Rajasthan Royals	70*	44	6	4	159.09
May 14 2011	Kolkata Knightriders	38	12	6	2	316.66	His innings: 4 4 . 4 4 . 6 6 2 4 4 out
May 17 2011	Kings XI Punjab	0	7	0	0	0	faced maiden over to Praveen Kumar
May 22 2011	Chennai Superkings	75*	50	4	6	150
May 24 2011	Chennai Superkings	8	9	0	1	88.88
May 27 2011	Mumbai Indians	89	47	9	5	189.36	opening stand 113 in 10.4 overs
May 28 2011	Chennai Superkings	0	3	0	0	0

In this series, if Gayle scored more than 8, Bangalore won. 🙂

3) Champions League 2011-12 (played for Royal Challengers Bangalore)
second highest run scorer: 6 innings, 257 runs, Ave 42.83, SR 178.47, 2 50s, 2 100s, 15 4s, 24 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Sep 23 2011	Warriors	23	14	2	2	164.28
Sep 29 2011	Kolkata	25	16	0	3	156.25
Oct 3 2011	Somerset	84	48	4	8	186.95
Oct 5 2011	SA	26	15	0	3	173.33	team chased down 215 – won by 2 wickets with 6 off last ball
Oct 7 2011	NSW	92	41	8	8	224.39	team chased down 204 in 18.3, 50 scored off 20 balls
Oct 9 2011	Mumbai	5	12	1	0	41.66

4) Stanbic Bank 20 Series in Zimbabwe (played for Matabeleland Tuskers)
highest run scorer (next best 225): 6 innings, 293 runs, Ave 58.60, SR 151.03, 1 100, 2 50s, 19 4s, 19 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Nov 25 2011	Mashonaland Eagles	27	31	3	0	87.09
Nov 26 2011	Mountaineers	61	38	3	5	160.52
Nov 27 2011	Southern Rocks	0	1	0	0	0
Nov 29 2011	Mid West Rhinos	109*	59	7	8	184.74	lost by 7 wickets (3 balls remaining)
Dec 2 2011	Mountaineers	45	34	3	2	132.25
Dec 3 2011	Mashonaland Eagles	51	31	3	4	164.51

5) Big Bash League in Australia, 2011-12 (played for Sydney Thunder)
8th highest run scorer: 7 innings, 252 runs, Ave 42.00, SR 150.00, 1 100, 2 50s, 11 4s, 22 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Dec 17 2011	Melb Stars	4	4	1	0	100
Dec 23 2011	Adelaide Strikers	100*	54	3	11	185.18	sixes 7 to 10 were in consecutive deliveries faced 100 off 53
Dec 30 2011	Melb Renegades	75	54	3	5	138.88
Jan 1 2012	Hobart Hurricanes	53	33	2	5	160.6
Jan 8 2012	Sydney Sixers	0	6	0	0	0
Jan 11 2012	Perth Scorchers	20	16	2	1	125
Jan 17 2012	Brisbane Heat	0	1	0	0	0

6) Bangladesh Premier League (played for Barisal Burners)
9th highest run scorer (when others played up to 12 innings): 5 innings, 288 runs, Ave 96.00, SR 187.01, 2 100s, 19 4s, 26 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Feb 10 2012	Sylhet Royals	101*	44	7	10	229.54
Feb 11 2012	Duronto Rajshahi	39	23	4	2	169.56
Feb 13 2012	Khulna Royal Bengals	2	10	0	0	20
Feb 14 2012	Dhaka Gladiators	116	61	6	11	190.16	lost by 21 runs, scored 116 out of 163 for team when dismissed
Feb 16 2012	Chittagong Kings	30*	16	2	3	187.5	retired hurt with groin injury

7) IPL 2012 in India (played for Royal Challengers Bangalore)
leading run scorer (next best: 590 runs): 14 innings, 733 runs, Ave 61.08, SR 160.74, 1 100, 7 50s, 46 4s, 59 6s

Date	Opponent	Runs	Balls	4s	6s	SR	Notes
Apr 10 2012	Kolkata Knightriders	2	8	0	0	25
Apr 12 2012	Chennai Superkings	68	35	2	6	194.28
Apr 15 2012	Rajasthan Royals	8	9	0	1	88.88
Apr 17 2012	Pune Warriors	81	48	4	8	168.75
Apr 20 2012	Kings XI Punjab	87	56	8	4	155.35
Apr 23 2012	Rajasthan Royals	4	8	0	0	50
Apr 28 2012	Kolkata Knightriders	86	58	7	6	148.27	was 38 off 38 balls at one stage
May 2 2012	Kings XI Punjab	71	42	6	4	169.04
May 6 2012	Deccan Chargers	26	22	1	2	118.18
May 9 2012	Mumbai Indians	82*	59	5	6	138.98
May 11 2012	Pune Warriors	57	31	3	6	183.87	50 off 24 balls
May 14 2012	Mumbai Indians	6	8	0	1	75
May 17 2012	Delhi Daredevils	128*	62	7	13	206.45	second highest T20 partnership of 204 with Kohli, team score of 215, 129 off the last 8.2 overs Gayle scored 90 off his last 28 balls faced
May 20 2012	Deccan Chargers	27	10	3	2	270

Aggregated, the seven series make stunning reading:

55 innings, 2613 runs, Ave 56.80, SR 169.46, 7 100s, 18 50s, 184 4s, 207 6s

Data courtesy of ESPN Cricinfo.

Chaitanya's Random Pages

June 26, 2012

Exports and Imports by Country

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June 23, 2012

The shortest day of the year

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June 11, 2012

Curious high or low temperature spots

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June 10, 2012

Outline proof of the extreme value theorem in statistics

Reference

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June 7, 2012

Chris Gayle in recent T20s

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