Some stuff » curve

whence the sun rises

admin — Fri, 27 May 2011 02:15:23 +0000

Somebody was commenting here that they thought the sun would rise from the Boston (south) side of the river in the morning, but in fact it rises from the Cambridge (north) side these days. A little more than ten years ago, sitting in my northern-latitude abode watching the sun set into the northwest, I wondered the same confused thing: why does the sun appear to venture into the northern part of the sky? (For reference, the sun’s direct projection on the earth never crosses north of the tropic of cancer, and that is south of here.)

The answer is recognized with the aid of this map, which shows the interesting effect due to the earth’s tilt and its spherical geometry. At every point along the day-night boundary curve, the sun’s rays are orthogonal to the curve and on the plane tangent to the earth’s surface at that point. In other words, the sun appears to rise (set) from (into) the direction orthogonal to the day-night boundary. This means between the spring and autumn equinoxes, the sun rises out of the northeast and sets into the northwest, and between the autumn and spring equinoxes, the sun rises out of the southeast and sets into the southwest.

This reminds me of a rather surprising method for direction-finding without a compass. It goes like this: point a stick in the direction of the sun, at any time during the day, and wait a short moment, then the direction of the new (very short, infinitesimal) shadow from the base of the stick points to the east. Here’s a short “proof”: Let the sun’s original location be called Q, let the base of the stick be called A and let the tip of stick be called B. So the vectors AB and AQ are collinear. At any given moment, the earth’s surface instantaneously rotates into the eastern direction at A, so in the earth’s local reference frame (the one we’re interested in) the sun’s apparent location moves slightly to the west to a new point Q’. Now, the shadow cast by the stick has a tip at a new point C (on the ground). Since A, B, C, Q, Q’ all must be coplanar, and the vector AC has no elevation component (same as QQ’), then the vector AC must be parallel to QQ’, and BQQ’ and BAC being similar triangles, AC must point to the east.

The trick is, you can’t wait too long, or else all the linear approximations and Euclidean geometry break down and you won’t get the right direction.

yield curve based on yield curve

admin — Tue, 23 Feb 2010 06:51:50 +0000

It occured to me the other day that based on certain models of the yield curve, it should make predictions about itself. It would be interesting to know how often it has been correct though.

As a start, we can convert the daily yield curve into implied short-term rates in the future. For instance, using the treasury yields as published here yesterday, we get these implied average rates over certain durations:

It says that the short-term rates will rise at about 1% per year and in five years it will stabilize at around 5%. Here is a close up at the short end.

And these translate into the following “standard” yield curves at 6 months (blue) and 1 year (green) from now, compared to currently (red).

There is no guarantee that this will be correct (that is something that needs to be checked against past data), but I distinctly recall that within the past two years the yield curves have, as a rule, been overly optimistic (too steep) on the short end.

uniform by three

admin — Sat, 22 Nov 2008 08:08:13 +0000

Here is a problem recently described to me. Apparently there is a more elegant solution (which may give more insight), but I don’t see it yet.

The problem: \(X, Y, Z\) are independent random variables uniformly distributed over [0,1]. What is the distribution of \((XY)^Z\)?

The conventional solution is to find the distribution of \(XY\) first, then of \((XY)^Z\).

The distribution of \(XY\) can be derived from its CDF \(F_{XY}(xy \le k)\), which is the total shaded area shown. The red curve is the function \(y=\frac{k}{x}\). This area is thus given by:

\(\int_k^1 \frac{k}{x} dx + k = k \ln(x) \vert_k^1 + k = -k \ln(k) + k\), for \(k>0\).

The PDF is the derivative of the above:

\(f_{XY}(k) = -\ln(k)\), for \(k>0\). The density is not well defined at \(k=0\).

The second part is to find the distribution in question. Here, the red curve is the function \(z=\frac{\ln(k)}{\ln(xy)}\). The CDF \(F_{(XY)^Z}((xy)^z \le k)\) is the total area to the left of the red curve. A column slice shaded in blue has probability per unit of \(z\) as labeled. Thus, the CDF is:

\(\int_0^k f_{XY}(v) dv (1-\frac{\ln(k)}{\ln(v)}) = \int_0^k -\ln(v) dv (1-\frac{\ln(k)}{\ln(v)})\)
\( = \int_0^k -\ln(v) dv + \int_0^k \ln(k) dv\)
\( = -v \ln(v) + v \vert_{\downarrow 0}^k + k \ln(k) = -k \ln(k) + k + k \ln(k) = k\), for \(k>0\).

For \(k=0\), we can fill in 0, because the minimum value of \((XY)^Z\) is 0, so it is the minimum of the support of its CDF.

So amazingly, \((XY)^Z\) has the CDF of (and is) a uniform distribution. How does that happen? What’s the intuition?