Wikipedia:Reference desk/Archives/Mathematics/2011 March 15

= March 15 =

Martingales & Optional Sampling Theorem
Hello everyone,

I was hoping you could help me out with a question on martingales and the optional sampling theorem. The problem is as follows:

Suppose $$a 0$$, find an appropriate martingale and use the optional stopping theorem to prove

$$\mathbb{E}\!\left[\exp\left({-\frac{1}{2}\beta ^2 \tau}\right)|B_0 = x\right] = \frac{\cosh{(\beta(x-(a+b)/2)}}{\cosh{(\beta(b-a)/2)}}$$.

Now first of all I'm having some trouble finding an appropriate martingale to apply to the question, could anyone offer a little help with this problem? Thankyou! :) Typeships17 (talk) 00:49, 15 March 2011 (UTC)


 * Could you explain your notation? For example, what is Bt? — Fly by Night  ( talk )  02:46, 15 March 2011 (UTC)


 * Have you tried the geometric Brownian motion $$e^{\beta B_t - \frac{1}{2}\beta^2 t}$$? This is a martingale.  Some variant of this should work for the problem.  Sławomir Biały  (talk) 03:52, 15 March 2011 (UTC)
 * I think Bt is the position of a Wiener process at time t. -- Meni Rosenfeld (talk) 09:59, 15 March 2011 (UTC)
 * Yes, sorry, i forgot to say Bt is a Brownian motion (or Wiener process). Hopefully the rest is self explanatory. And thanks, I'll take a look at the geometric brownian motion, the left hand side would then arise pretty obviously, though where the right hand side comes from is currently beyond me, I'll see if I can figure anything out. Typeships17 (talk) 17:08, 15 March 2011 (UTC)
 * Let us know how it goes, either way. ;-) Sławomir Biały  (talk) 12:36, 15 March 2011 (UTC)
 * Wait, if Bt is a Brownian motion, surely we can't set B0=x? I was sure it was meant to be a BM, in keeping with the notation I've seen previously, but perhaps I'm just confused... Typeships17 (talk) 17:14, 15 March 2011 (UTC)
 * There's no significant loss in generality in taking x=0. Then it's the usual Brownian motion starting at the origin, if it helps. Sławomir Biały  (talk) 17:55, 15 March 2011 (UTC)
 * Okay, so we want to say WLOG Bt is a Brownian motion, and prove $$\mathbb{E}\!\left[\exp\left({-\frac{1}{2}\beta ^2 \tau}\right)\right] = \frac{\cosh{(\beta(a+b)/2}}{\cosh{(\beta(b-a)/2)}}$$. The left hand side is clearly $$\mathbb{E}\left[ M_0\right]$$, where $$M_t=e^{\beta B_t - \frac{1}{2}\beta^2 t}$$, and the optional sampling theorem says that equals $$\mathbb{E}\left[ M_{\tau}\right]$$ - but how do I evaluate $$\mathbb{E}\left[ M_{\tau} \right]$$? I'm not totally sure how to take the expectation when the step number is a stopping time rather than a 'fixed value' - do we use the iterated expectation formula? Advice would be much appreciated! Typeships17 (talk) 18:24, 15 March 2011 (UTC)

Consider the martingale
 * $$X_t = e^{-\frac{1}{2}\beta^2t}\cosh(\beta (B_t - (a+b)/2)).$$

Evaluate $$\mathbb{E}^x[X_\tau]$$ in two different ways. First, by optional sampling,
 * $$\mathbb{E}^x[X_\tau] = X_0 = \cosh(\beta(x-(a+b)/2))$$

Second, since at $$t=\tau$$, $$\cosh(\beta (B_\tau - (a+b)/2))=\cosh(\beta(a-b)/2)$$ is deterministic,
 * $$\mathbb{E}^x[X_\tau] = \mathbb{E}^x\left[e^{-\frac{1}{2}\beta^2\tau}\right]\,\cosh(\beta (a-b)/2)$$

as desired. Sławomir Biały (talk) 00:16, 16 March 2011 (UTC)
 * That's brilliant, thank you Sławomir! You know I don't think I've ever really seen an argument like that before, it's very nice. Thanks again :-) Typeships17 (talk) 10:25, 16 March 2011 (UTC)

Singular value decomposition
Just checking that the article singular value decomposition is correct. Given M is an mxn matrix and M=UΣVT, it claims U is an mxm matrix. However, examples I've seen make U an mxn matrix, Σ an nxn matrix, and V an nxn matrix. It is possible that the article has switched the meaning of U and Σ compared to other articles I've read. -- k a i n a w &trade; 04:14, 15 March 2011 (UTC)


 * The article makes sense to me. In particular, U is supposed to be an unitary matrix, and for that it has to be square. –Henning Makholm (talk) 04:42, 15 March 2011 (UTC)
 * See Singular_value_decomposition. The "vanilla" SVD indeed has square U and V matrices, but to save space you can drop unnecessary rows and columns. The "Thin SVD" is the form you've seen. -- Meni Rosenfeld (talk) 09:43, 15 March 2011 (UTC)


 * Thanks. That explains it perfectly. -- k a i n a w &trade; 12:27, 15 March 2011 (UTC)

Latex Pictures
I want to put some large images into my latex file.

However because of the margins of my page, the pictures get pushed to the right of the page. I cannot make them centered unless I make them very small.

I am using the graphicx package, as follows:

\begin{figure}[H] \begin{center} \includegraphics[scale=0.4]{picture.png} \caption{etc} \end{center} \end{figure}

However the "center" isn't actually centering it, due to the fatness of the margin on the left.

How do I get the picture to be actually centered, overriding the margin? Like choosing "in front of text" when you're putting a picture into microsoft word? 130.102.158.15 (talk) 05:31, 15 March 2011 (UTC)


 * Hmm, in plain TeX I would just add \hss on either side of the over-wide box, but that doesn't seem to work in LaTeX. Perhaps one of the suggestions in this TeXblog post will work for you. —Bkell (talk) 10:28, 15 March 2011 (UTC)

Expression for partial sums of sequence
Is there a closed-form expression for the nth term of the sequence


 * 1, 3, 5, 8, 11, 14, 18, 22, 26, 30, ...

which are the partial sums of the sequence


 * 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...

containing one 1, two 2s, three 3s etc. ? The asymptotic behaviour of the nth term (of the first sequence) is
 * $$\frac{(2n)^{\frac{3}{2}}}{3}$$

but I am wondering if an exact expression is known. Gandalf61 (talk) 14:41, 15 March 2011 (UTC)


 * The k(k+1)/2 term is k(k+1)(2k+1)/6 from the formula for the sum of squares, so you need to find the next k down that gives your n and add up from there


 * k(k+1)/2 = n would give k = (-1+sqrt(1+8k))/2 so we choose the next integer below this as k. That will give a sum for all the terms up to the k(k+1)/2 one and we just add (n-k/k(+1)/2)k to the sum. Dmcq (talk) 15:11, 15 March 2011 (UTC)


 * Okay, so putting that all together I get:
 * $$n(k+1)-\frac{k(k+1)(k+2)}{6} \text { where } k=\left\lfloor\frac{\sqrt{8n+1}-1}{2}\right\rfloor$$
 * Not pretty, but it works. Thank you. Gandalf61 (talk) 16:00, 15 March 2011 (UTC)
 * You can also write in a slightly less not-pretty way as
 * $$nk-\frac{k^3-k}{6} \text { where } k=\left\lfloor\frac{\sqrt{8n+1}+1}{2}\right\rfloor$$
 * -- Meni Rosenfeld (talk) 16:13, 15 March 2011 (UTC)
 * That's positively beautiful compared to what I expected :) Dmcq (talk) 18:54, 15 March 2011 (UTC)

Calculator
How do you get the "exact" result on your calculator? (for example: square root 2/2 instead of 0.707106781) Thanks. — Preceding unsigned comment added by Atacamadesert12 (talk • contribs) 16:46, 15 March 2011 (UTC)
 * What kind of calulator do you have? Most fancy (graphing) calculators can convert rational decimals to fractions pretty well, but most can't do the kind of thing you suggest with irrational numbers. (I don't know of any ordinary handheld calculator that can do that.) Staecker (talk) 17:35, 15 March 2011 (UTC)
 * By the way, you can do this sort of thing at Plouffe's Inverter, which for your example gives various different forms, some more useful than others. Staecker (talk) 17:43, 15 March 2011 (UTC)

I have Sharp EL-520W — Preceding unsigned comment added by Atacamadesert12 (talk • contribs) 18:17, 15 March 2011 (UTC)
 * I have no knowledge about this calculator, but if you want to do computations and get an exact result, you can use Wolfram|Alpha (example). -- Meni Rosenfeld (talk) 19:43, 15 March 2011 (UTC)

You are asking about doing symbolic algebra on your calculator. There have been some calculators that can do it, but these days it probably makes more sense to use a computer, either with software like Sage or (if you must) with a web service like Wolfram Alpha. 75.57.242.120 (talk) 22:13, 15 March 2011 (UTC)

powerball jackpot
What it the statistical odds that the next powerball jackpot will exceed $500 mil before being won? —Preceding unsigned comment added by 75.170.239.188 (talk) 17:08, 15 March 2011 (UTC)


 * It's going to be very difficult to calculate, as the size of the jackpot depends on how many tickets were sold, as well as how many times it rolls over (related to the number of tickets sold). The number of tickets sold varies, and depends somewhat on the size of the jackpot (larger jackpots mean more tickets sold). You really can't make the calculation without specifying some model of ticket sales. All that said, the largest Powerball jackpot so far was $365 million, so the chance that it'll hit $500 million on the next jackpot win is for all practical purposes indistinguishable from zero. -- 140.142.20.229 (talk) 22:51, 15 March 2011 (UTC)


 * It shouldn't be too difficult. Assuming a random distribution, the number of tickets sold is proportional to the chances of it being a roll-over. — Fly by Night  ( talk )  23:29, 15 March 2011 (UTC)

Continuous compound interest
Every calculus book I read has examples on "continuously compounded" interest. Does any financial instrument in the real world behave this way? My bank account sure doesn't. Our sections on continuous compounding don't seem to give any real examples. Staecker (talk) 17:38, 15 March 2011 (UTC)


 * I doubt there would be any financial institutions that do. The more often interest is calculated, the better it is for the customer. But I'm sure that there are examples of continuous exponential growth in nature. — Fly by Night  ( talk )  18:13, 15 March 2011 (UTC)


 * I know of at least one (funny) example. A bank was offering continuous compounding on it's savings account, and had a TV ad featuring an exhausted accountant who was recalculating the interest on your account "continuously", never sleeping.  This may have just been a gimmick by one bank, though.


 * In general, I don't see the advantage. If the banks wants to give a higher interest rate, just offer a higher periodic interest rate instead.  Using complex math that the majority of their customers can't understand is just going to cause confusion and frustration among the customer base and the tellers who are asked to explain it all.


 * Also, if the bank ends up paying more when offering 10% interest compounded continuously, than 10% interest compounded monthly, but the customer is unaware of the difference, it's not good for the bank. They may also include the "effective annual yield", but skeptical customers may ignore that and use the lower nominal rate to choose their bank.


 * Now, in the cases of businesses that charge you interest, I would think that continuous interest would be a way to squeeze a bit more out of the customers without appearing to do so. However, the argument of not wanting to alienate customers incapable of understanding the calculations still applies. StuRat (talk) 20:55, 15 March 2011 (UTC)


 * I'd hardly call it "complex math". The formula is A(t) = Pert where A(t) is the amount in the bank after time t, P is the principal investment and r is the annual interest rate. — Fly by Night  ( talk )  21:10, 15 March 2011 (UTC)


 * Perhaps not complex for the average person here, but it is for the average bank customer. They'd just think the bank was making that formula up and want to know where the heck "e" came from. StuRat (talk) 07:42, 17 March 2011 (UTC)

I assume that no ordinary bank accounts offer this- I was wondering about more complicated financial instruments. As far as I understand, banks like to manage accounts according to occasional "transactions", each of which changes the balance. My bank deposits my interest every time they compound it in a transaction. Continuous compounding would mess up this whole system. But I wonder if there are some esoteric financial gadgets not typically owned by ordinary people that work by continuous compounding. There are certainly plenty of crazier and much more complicated things out there. Staecker (talk) 22:33, 15 March 2011 (UTC)


 * Apparently credit card interest is typically compounded daily. That's pretty close to continuous.  Presumably there are practical constraints on having instruments compounded more often than daily.   Sławomir Biały  (talk) 23:07, 15 March 2011 (UTC)


 * There's a patent on sub-day compounding, but really continuous compounding may still be free ... –Henning Makholm (talk) 23:45, 15 March 2011 (UTC)

They're not financial instruments per se, but models for valuing stock options generally treat the risk-free interest rate as continuously compounded. The assumption makes sense in the context of the model, since the point is to arrive at an option's value rather than calculate the interest at a particular point in time. —Preceding unsigned comment added by 12.186.80.1 (talk) 20:28, 17 March 2011 (UTC)
 * Thanks- that's a good answer. Staecker (talk) 22:45, 17 March 2011 (UTC)

Bayesian probability
Are there any introductory probability texts that give solely Bayesian probability without providing or requiring a background in classical probability? (And obviously if so, what are some) Thanks. 72.128.95.0 (talk) 23:31, 15 March 2011 (UTC)


 * Jaynes' book is probably what you want. I haven't read it myself though. -- Meni Rosenfeld (talk) 09:06, 16 March 2011 (UTC)

The point where "classical probability theory" starts to differ from "Bayesian probability thoery" is when you get into applications, and at that point "classical probability theory" ceases to consist only of probability theory, whereas the point of Bayesianism is that it's all to be reduced to probability theory. So as long as it's only probability theory, it's consistent with both Bayesianism and frequentism. Michael Hardy (talk) 16:04, 16 March 2011 (UTC)
 * Is there a straightforward example application where the Bayesian and frequentist approaches give differing answers? 75.57.242.120 (talk) 10:35, 18 March 2011 (UTC)
 * There's no shortage of papers that debunk frequentism and give examples. Here's an example that came up in a search: (I haven't read it). -- Meni Rosenfeld (talk) 11:44, 18 March 2011 (UTC)