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April 6

Sums using multiple variables: A question about sigma notation

Hello, I have a question about sigma notation. One could write the following, having the understood meaning shown afterward in an equality: $\sum _{i=1}^{4}{\binom {7}{i}}={\binom {7}{1}}+{\binom {7}{2}}+{\binom {7}{3}}+{\binom {7}{4}}$

However, what if we have two variables, i and j, with i going UP from 1 to 3 and j going down from 4 to 2, simultaneously? (NOT a double sigma where all possibilities of i,j are summed!) I want to express the following sum: ${\binom {1}{4}}+{\binom {2}{3}}+{\binom {3}{2}}$

I am tempted to write as follows, or similar: $\sum _{i=1,j=4}^{i=3,j=2}{\binom {i}{j}}$ OR $\sum _{i=1,j=4}^{3,2}{\binom {i}{j}}$ etc...

... but this seems silly! How can i PROPERLY express what i am trying to, using sigma notation? (Or a better way, if what im trying to do as short hand isnt appropriate)

216.173.144.188 (talk) 04:41, 6 April 2016 (UTC)[reply]

One option is

\sum _{i=1}^{3}{\binom {i}{5-i}}

. -- ToE 06:04, 6 April 2016 (UTC)[reply]

Yes, I think that's the way it would usually be done. If both i and j were used a lot of times in the expression, I suppose you could write something like

\sum _{i=1}^{3}{\binom {i}{j}}

where

j=5-i

and it would be understood. --69.159.61.172 (talk) 06:47, 6 April 2016 (UTC)[reply]

Okay, thank you, this makes sense! However, it is a simplification based on what i guess now i see as a poor substitute example. Let me ask about my actual situation and see what the correct way is to write this concisely as a mathematician. We have the following:

Given $n\in \mathbb {N} ^{+}$ , Consider $\sum _{j=\lfloor {\frac {n}{2}}\rfloor }^{0}{\binom {n-j}{j}}$ ...

Does this make sense? Is there any reason why the index of a sum cant decrease instead of increase?

(It took me a while to realize i could express i in terms of n and j together, which helped in this)

216.173.144.188 (talk) 13:20, 6 April 2016 (UTC)[reply]

In the sigma notation the index can't decrease because Summation#Formal_definition says that

\sum _{i=a}^{b}g(i)=0

\,

, for b < a. The fix is simple though, just write

\sum _{j=0}^{\lfloor {\frac {n}{2}}\rfloor }{\binom {n-j}{j}}

. Egnau (talk) 16:02, 6 April 2016 (UTC)[reply]

Indeed,

Thank you all very much for your input. I consider this matter closed. 216.173.144.188 (talk) 16:35, 6 April 2016 (UTC)[reply]

The Iverson bracket is useful.

\sum _{i=1}^{3}\sum _{j=0}^{\infty }[i+j=5]{\binom {i}{j}}

Bo Jacoby (talk) 22:31, 6 April 2016 (UTC).[reply]

What are the odds that a person will die on his birthday?

What are the odds that a person will die on his birthday? Is it 1 out of 365? Or is it more complicated than that? I was surprised to see that this happened with Merle Haggard today. Thanks. Joseph A. Spadaro (talk) 19:49, 6 April 2016 (UTC)[reply]

yes, would have to be, if looking at whole population...or whatever it end up being if include the extra leap days....UNLESS it can be determined that more people tend to be born in the spring or summer and more people tend to die in the fall or winter.....68.48.241.158 (talk) 20:26, 6 April 2016 (UTC)[reply]

But once you have a birthday, you have a birthday. That's that. So, it's simply a matter of "matching" the death day. No? Joseph A. Spadaro (talk) 20:28, 6 April 2016 (UTC)[reply]

Close to 365, unless they were born on leap day. I suppose a surprise and big birthday meal might put the death rate ever so slightly higher that day. StuRat (talk) 20:27, 6 April 2016 (UTC)[reply]

if 65% of population is born in spring or summer and 65% of population dies in fall or winter, then looking at the whole population...a human being has a smaller chance than one 1/365(or so) of dying on her birthday... 68.48.241.158 (talk) 20:34, 6 April 2016 (UTC)[reply]

I am asking about a specific person, not just any generic person. And a specific person, once born, already has a set birth date (whatever that date might be). So, the question is not so much: "What are the odds that any one of us human beings will die on our birthday?" The question is more like: "What are the odds that Brad Pitt will die on his birthday?" Which is another way of saying: "What are the odds that Brad Pitt will die on December 18?". Joseph A. Spadaro (talk) 21:14, 6 April 2016 (UTC)[reply]

if asking specific person than depends on individual factors..so unanswerable...person might be dying of cancer, only have 6 weeks to live...birthday 6 months away.....?????68.48.241.158 (talk) 21:21, 6 April 2016 (UTC)[reply]

but if brad pitt is expected to live many, many more years (and human being are more likely to die in fall or winter) then he has a smaller chance of dying on his birthday....68.48.241.158 (talk) 21:23, 6 April 2016 (UTC)[reply]

... if he is living in the southern hemisphere? Dbfirs 21:29, 6 April 2016 (UTC)[reply]

^right, my mistake...general reasoning okay though, I think...68.48.241.158 (talk) 21:35, 6 April 2016 (UTC)[reply]

Actually, is there a mistake? December 18 is still technically fall, I believe (not winter). I think winter starts on December 21 or so. No? Joseph A. Spadaro (talk) 22:29, 6 April 2016 (UTC)[reply]

There are many definitions of when seasons start. Don't be fooled by news people (especially in North America) who like to talk about "the official first day of winter". Besides, everyone knows the official first day of winter is in October. --69.159.61.172 (talk) 23:07, 6 April 2016 (UTC)[reply]

Summer starts on December 1st in almost all of the temperate southern hemisphere, though I expect someone will be able to find an odd exception. Dbfirs 07:46, 7 April 2016 (UTC)[reply]

Agreed. There are slightly more deaths in winter where I live and also where you live, but that will not be the case in some other parts of the world. Dbfirs 21:50, 6 April 2016 (UTC)[reply]

That would be true if seasonal variation were as great in humans as it is in some other animals. The answer will be approximately "364:1 against" (the question asked for odds, not probability), but with slight variations as mentioned above. Dbfirs 21:06, 6 April 2016 (UTC)[reply]

certainly there's some variation so determining the actual odds or probability (is there a difference in this context?) would in practice be complicated...68.48.241.158 (talk) 21:14, 6 April 2016 (UTC)[reply]

Yes, I wasn't disagreeing, just commenting that the effect would be small. It would also vary by region and culture. I agree that the fine detail would be complicated to calculate. In the future, it might be possible to predict, from an individual's DNA, address, and lifestyle, the most likely time of year of death, but the probability is unlikely to change significantly from 1/365 (the small correction for leap years possibly cancels out the small birthday surprise effect or birthday parachute jump). The answer for Brad Pitt will be very close to the answer for any other individual. Dbfirs 21:26, 6 April 2016 (UTC)[reply]

(Not to be morbid, but it's a morbid question anyway) Infant mortality may dominate other effects. 11000 / 4000000 ≈ 1/350 of all babies born in the U.S. die within 24 hours of birth. Probably more than 1/700 die on the same date (more births are in the morning, and probably more deaths are earlier in the 24-hour period too). That's quite large compared to 1/365. -- BenRG (talk) 04:55, 7 April 2016 (UTC)[reply]

Good point. We should probably restate the problem as "What is the probability of dying on your birthday, but not your birth date". StuRat (talk) 05:17, 7 April 2016 (UTC)[reply]

In that case, mortality is highest on the day after the birthday, and still high the days after that, from infant deaths with a small delay after birth. Then, deaths on the birthday itself are probably even lower than average since it is the last date to appear after we start counting, and therefore least affected by infant mortality. Unless some other intrinsic birthday effect offsets this, of course. If all effects related to infant mortality are to be excluded from the question, one should exclude all data below age 5 or so. Gap9551 (talk) 16:18, 8 April 2016 (UTC)[reply]

Yes, I apologise to all new-born babies for not considering them as "person"s. Dbfirs 07:56, 7 April 2016 (UTC)[reply]

Check out "Variation of mortality rate during the individual annual cycle", which investigates all births/deaths in the city of Kiev over the period 1990-2000. Kiev men showed a 44.4% excess in deaths on their birthday compared to the expected value (women slightly less at 36.2% excess), and it looks like other studies have found similar results. So working out the exact odds would take a lot of actuary work, but it looks like it's something more like 1:300 than 1:365. Smurrayinchester 08:46, 7 April 2016 (UTC)[reply]

I've started a page Birthday effect which collects the studies. Please feel free to improve it. Smurrayinchester 11:53, 7 April 2016 (UTC)[reply]

Thank you for that excellent summary of the research. The effect is much bigger than I expected. Dbfirs 14:14, 7 April 2016 (UTC)[reply]

Very interesting. I will have to read that new article. Thanks. Joseph A. Spadaro (talk) 18:04, 7 April 2016 (UTC)[reply]

I downloaded the first third of the Social Security Death Index (2.9 GB). From the 28,607,398 recorded birth / death pairs, I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). It should be noted that the SSDI only records deaths of individuals who were issued US social security numbers. As SSNs are not issued to individuals who die prior to the birth being officially registered, none of the records include infants who died on the day they were born. Dragons flight (talk) 18:50, 7 April 2016 (UTC)[reply]

As for seasonal differences:

	January	February	March	April	May	June	July	August	September	October	November	December
Births	2420302	2269663	2525845	2330376	2386245	2296193	2442604	2487487	2457619	2387806	2226876	2376382
Deaths	2708214	2407497	2539914	2348925	2331869	2217327	2290327	2248835	2197259	2391209	2362526	2563496

Dragons flight (talk) 18:59, 7 April 2016 (UTC)[reply]

Same dataset has 2,393,596 individuals (8.37%) dying in the same month that they were born, or a chance of 1 in 11.952. Dragons flight (talk) 19:10, 7 April 2016 (UTC)[reply]

Why would there be such a disparity between what we would expect (statistically) and the actual results? That does not seem to make sense. I am referring to this comment: I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). Thanks. Joseph A. Spadaro (talk) 21:00, 7 April 2016 (UTC)[reply]

What is the distribution of death dates? E.g., how many days (out of 365) have frequency less than 0.16% ? --JBL (talk) 21:09, 7 April 2016 (UTC)[reply]

Hhmmm, I've now taken a closer look at the data and it appears that about half of the SSDI death dates are recorded without a real day of death, e.g. March 0, 1979. Some of the birth dates also are also missing detail (but a much smaller fraction than the death dates). If I exclude all of the birth / death pairs that don't involve fully-formed dates, I am left with 14,879,058 individuals. Of these 44,665 were recorded as dying on their birthday (0.3002%; a chance of 1 in 330). That is now slightly more than the expected 1 in 365, though much closer than the initial estimate. Dragons flight (talk) 07:19, 8 April 2016 (UTC)[reply]

Looking a bit further:

Birthday + X days	Chance of dying (according to SSDI records)
-40 days	1 in 364
-30 days	1 in 366
-20 days	1 in 366
-10 days	1 in 366
-5 days	1 in 366
-4 days	1 in 366
-3 days	1 in 363
-2 days	1 in 370
-1 days	1 in 365
Birthday	1 in 330
+1 days	1 in 361
+2 days	1 in 361
+3 days	1 in 363
+4 days	1 in 364
+5 days	1 in 365
+10 days	1 in 363
+20 days	1 in 363
+30 days	1 in 365
+40 days	1 in 366

Dragons flight (talk) 08:08, 8 April 2016 (UTC)[reply]

that seems more in line with common sense...the previous estimate seemed weird..68.48.241.158 (talk) 11:54, 8 April 2016 (UTC)[reply]

confused about the chart though...saying only 1/364 chance dying the time frame of 40 days before birthday?? that can't be right...??68.48.241.158 (talk) 16:33, 8 April 2016 (UTC)[reply]

It's the chance of dying exactly that many days before or after your birthday. Dragons flight (talk) 17:00, 8 April 2016 (UTC)[reply]

That chart seems strange. All of the time intervals are listed with odds of very close to 1 in 365. They are all 362, 363, 364, 366, etc. Why would the exact birthday be the only one with a "weird" amount listed for odds as 330? That (the birthday) is the only one on that chart that really deviates from the expected 365 value. Why? Joseph A. Spadaro (talk) 18:25, 8 April 2016 (UTC)[reply]

Or does that strange value (330) include babies that die on the exact day they are born? But even that theory wouldn't make sense, I don't think? Joseph A. Spadaro (talk) 18:31, 8 April 2016 (UTC)[reply]

good question, as he said infants who died day they were born are not included...this would seem the only possibility that skew the numbers that drastically....68.48.241.158 (talk) 18:41, 8 April 2016 (UTC)[reply]

Yes. But doesn't the concept of "infant mortality" extend to the 1, 2, 3, 4, 5 or so days after the birth, also? Not just the exact birthday. So, those other surrounding days should also end up being skewed. No? Joseph A. Spadaro (talk) 19:13, 8 April 2016 (UTC)[reply]

Perhaps you should go reread the post in this thread in which Dragons flight explains where the data comes from. --JBL (talk) 20:15, 8 April 2016 (UTC)[reply]

well, how you explain the huge anomaly on the actual birthday???68.48.241.158 (talk) 20:20, 8 April 2016 (UTC)[reply]

Exactly! Joseph A. Spadaro (talk) 20:40, 8 April 2016 (UTC)[reply]

For a specific person - and estimating the odds on a specific day, there are more things to consider. Imagine, for example, two people who were born just two days apart - with identical lifestyle, identical current health, etc. Suppose A's birthday was yesterday, and B's birthday is tomorrow. If (for example) they both have some terminal disease with less than a month to live - then B can still die on his birthday - but A cannot. If they'll both live for less than two years, then B has two chances to die on his birthday - but A has only one. Of course if we do the estimation two days from now, their odds of having birthday deaths are very similar again. SteveBaker (talk) 18:55, 8 April 2016 (UTC)[reply]

not really relevant...yes, if looking at one specific person all kinds of things could come in play...but this is looking at odds for a large population as a whole...(you're absolutely right about what you say though...this was mentioned earlier in thread too)..68.48.241.158 (talk) 19:07, 8 April 2016 (UTC)[reply]

A couple of other things to consider, if people can find stats for them: 1) Placebo effect - could potentially result in people close to death and close to their next birthday managing to hang on until they reach it. 2) People celebrating their birthday by getting really, really drunk, and dying of alcohol-related accident or illness. Iapetus (talk) 10:09, 11 April 2016 (UTC)[reply]

Thanks, all. Joseph A. Spadaro (talk) 03:50, 12 April 2016 (UTC)[reply]

Broken down over kinds of causes of death, it seems certain ones would appear close to randomly with 4/1441 the chance of occurrence on the person's birthday, while others such as motor vehicle accident, suicide, and accidental drug overdose having greater chance of happening on birthday.4.35.219.219 (talk) 00:27, 13 April 2016 (UTC)[reply]