Finance-Theory · Unit 22 · Video 1 · Interactive Practice

Market Efficiency: How Stocks Found the Challenger Culprit

IKey Formulas

Formula	Name	What it says
$R = \dfrac{P_t}{P_0} - 1$	Return over an interval	The later price divided by the earlier price, minus one; a fall over several intervals is measured from the first price, not by adding the interval returns
$\dfrac{P_t}{P_{\text{open}}}$	Price relative to the open	Every stock starts at $1$ , so differences in price level drop out and the paths can be compared
$\hat R_{\text{MT}} = \bar R_{\text{others}} \times \dfrac{s_{\text{MT}}}{\bar s_{\text{others}}}$	Exposure-scaled fall	The fall Thiokol would show if every contractor lost in proportion to its NASA share of sales $s$ : $2.62\% \times 18.23 / 10.45 \approx 4.6\%$ , against an actual $11.86\%$
$z = \dfrac{R - \bar R}{\sigma}$	Standard deviations from the average	Thiokol's one-day fall was $6.49$ standard deviations from its prior three-month average ( $\|z\| = 6.49$ ); each of the other three was under $2$

Key Insight: By the close on January 28, 1986, Morton Thiokol had fallen about four times as far as the other three contractors, with no announcement, no reported insider trades and no analyst concern. The Rogers Commission reached the same verdict on June 9, 1986: prices can aggregate dispersed information.

IIReturns Over the Recorded Intervals

Each interval's return is the later price over the earlier price, minus one, and only Thiokol has no noon price.

From To Firm in the formula

IIIThe Paths Separate Before the Close

Rescaled to start at $1$ , all four stocks drop at the crash; then only Morton Thiokol keeps sliding.

Time of day 4:00

💡 The market had no announcement to react to: the authors found no reported insider trades that day, several analysts called Thiokol's fall an overreaction, and it took the press about a week, until February 3, for the New York Times to report that Morton Thiokol was facing the closest scrutiny.

IVWas It Just NASA Exposure?

If each contractor fell in proportion to its NASA share of sales, Thiokol should have lost far less than it did.

VQuiz Questions

Problem 1 · Return Over One Interval

Given: An illustrative firm A trades at $24.00 just before a piece of bad news and at $22.80 half an hour later. Find its return over the half hour.

✅ Correct! The later price is

0.95

of the earlier one, so the stock lost exactly

5\%

of its value.

❌ Not quite. That divides the $1.20 drop by the later price, $22.80. A return is measured against the earlier price:

P_t / P_0 - 1

with

P_0

the price before the news.

❌ Check the sign. The price went down, so

P_t / P_0

is below

1

and the return is negative.

❌ Not quite. That is the dollar change per share, $1.20, written as a percent. The return divides the change by the starting price.

Show solution

With $P_0 = 24.00$ and $P_t = 22.80$ (dollars per share):

R = \frac{P_t}{P_0} - 1 = \frac{22.80}{24.00} - 1 = 0.95 - 1 = -0.05 = -5.00\%

The division is exact, so the return is exactly $-5\%$ .

Problem 2 · A Fall Across a Trading Halt

Given: An illustrative firm B traded at $28.00 just before bad news. Trading was halted; the first trade after the halt was at $26.32, and the next recorded trade was at $25.76. Find its return from the pre-news price to the last recorded trade.

✅ Correct! Measured from the pre-news price, $25.76 is

0.92

of $28.00: an

8\%

fall. The halt only delays when that fall shows up in a trade.

❌ Not quite. That is the return to the first trade after the halt only. The question asks for the fall to the last recorded trade, which went lower.

❌ Not quite. That is the return between the two post-halt trades,

25.76 / 26.32 - 1

. The fall from before the news starts at $28.00.

❌ Close, but… that adds the two interval returns,

-6.00\% + (-2.13\%)

. Returns compound rather than add: measure the whole fall directly as

P_t / P_0 - 1

from the first price.

Show solution

The return over the whole span uses only its two end prices:

R = \frac{25.76}{28.00} - 1 = 0.92 - 1 = -8.00\%

The interval returns chain multiplicatively: $\frac{26.32}{28.00} = 0.94$ and $\frac{25.76}{26.32} \approx 0.9787$ , and $0.94 \times \frac{25.76}{26.32} = \frac{25.76}{28.00} = 0.92$ exactly. Adding $-6.00\%$ and $-2.13\%$ overstates the fall slightly.

Problem 3 · How Much Does Exposure Explain?

Given: After an accident, three illustrative peer firms fell $3.0\%$ on average, and their average share of sales to the affected program was $12\%$ . Firm D sold $20\%$ of its output to the program and fell $8.5\%$ . Assume, as in the video's check, that the fall from exposure alone is proportional to the share of sales.

What fall does exposure alone predict for firm D?

What share of D's actual fall does that explain?

✅ Correct! Exposure predicts a

5.0\%

fall, about

59\%

of the

8.5\%

that happened. Unlike Thiokol, where exposure explained well under half, most of D's fall fits the exposure story; the market's extra judgment about D is the remaining

3.5

points.

❌ Not quite. That scales by

12/20

, the ratio upside down. D is more exposed than its peers, so its predicted fall must be larger than their

3.0\%

: multiply by

20/12

❌ Not quite. That is the peers' average fall with no adjustment. D sells a larger share to the program, so the proportional story predicts more:

3.0\% \times 20/12

❌ Not quite. That is D's actual fall. The question asks what exposure alone would predict, which comes from the peers' average fall scaled by relative exposure.

❌ Check the prediction. Scale the peers' average fall by D's share relative to theirs:

3.0\% \times 20/12

❌ Not quite. That is

3.0/8.5

, using the peers' unscaled fall. Use the exposure-scaled prediction,

5.0\%

, as the explained part.

❌ Not quite. That is

8.5/5.0

, the ratio upside down. The explained share is the predicted fall divided by the actual fall.

❌ Not quite. The predicted

5.0\%

is less than the actual

8.5\%

, so exposure leaves

3.5

points of the fall unexplained.

❌ Check the share. Divide the exposure-predicted fall by the actual fall.

Show solution

Step 1: Prediction from exposure. If falls are proportional to the share of sales, D's fall should be the peers' average scaled by D's share relative to theirs:

\hat R_D = 3.0\% \times \frac{20}{12} = 5.0\%

Step 2: Share explained.

\frac{5.0\%}{8.5\%} \approx 0.588 \approx 59\%

Exposure accounts for more than half of D's fall; the remaining $8.5 - 5.0 = 3.5$ points are what the market added beyond exposure. In the video's Challenger check the same method gave $2.62\% \times 18.23/10.45 \approx 4.6\%$ against an actual $11.86\%$ , well under half, which is why the remainder was read as the market's judgment of fault.

Problem 4 · Measuring a Fall in Standard Deviations

Given: Over the months before some bad news, illustrative firm F's daily returns averaged $+0.1\%$ with a standard deviation of $1.2\%$ ; illustrative firm G's also averaged $+0.1\%$ , with a standard deviation of $4.0\%$ . On the day of the news F fell $5.9\%$ and G fell $8.9\%$ . Find F's $z$ , then judge which fall stands further out from its own history.

What is F's $z$ for the day?

Which day's fall is more unusual for its own stock?

✅ Correct! F's fall is

5

standard deviations below its average day; G's larger fall is only

2.25

for a stock that swings far more. This is the comparison the video makes: Thiokol's fall was more than six standard deviations from its average, each other contractor's under two.

❌ Not quite. That is

-5.9 / 1.2

, which leaves out the average.

z

measures the distance from the average day: subtract the

+0.1\%

first.

❌ Not quite. That multiplies the gap by the standard deviation,

-5.9 \times 1.2

z

divides by it, to count how many standard deviations the day lies out.

❌ Check the sign. That adds the average,

(-5.9 + 0.1)/1.2

. The day lies below an average of

+0.1\%

, so the gap is

-5.9 - 0.1 = -6.0

points.

❌ Check the formula. Use

z = (R - \bar R)/\sigma

with F's own average and standard deviation.

❌ Not quite. G's daily returns swing by

4\%

as a matter of course, so an

8.9\%

fall is only

(-8.9 - 0.1)/4.0 = -2.25

standard deviations. The size of a move alone does not say how unusual it is.

❌ Not quite. Passing a threshold is not the same as being equally far out:

5

standard deviations is far rarer than

2.25

❌ Not quite. Dividing by each stock's own standard deviation is exactly what makes them comparable:

z

puts both falls on the same scale.

❌ Check the comparison. Compute G's

z

the same way and compare the two.

Show solution

Step 1: F's $z$ .

z_F = \frac{R - \bar R}{\sigma} = \frac{-5.9\% - 0.1\%}{1.2\%} = \frac{-6.0}{1.2} = -5.00

Step 2: G's $z$ .

z_G = \frac{-8.9\% - 0.1\%}{4.0\%} = \frac{-9.0}{4.0} = -2.25

G lost more in percent, but for a stock that volatile such a day is not far out of the ordinary. F's fall lies $5$ standard deviations below its average day, so it is the one that singles F out: the same reason the video compares Thiokol's fall with its own history, not only with the other contractors' falls.

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