Finance-Theory · Unit 15 · Video 4 · Interactive Practice

The Risk-Return Graph: Why Investors Head Northwest

IKey Formulas

Formula	Name	How to read it
$\text{asset } i \;\mapsto\; \big(SD[R_i],\ E[R_i]\big)$	A point on the graph	Risk across the horizontal axis, expected return up the vertical axis
$SD[R_A] < SD[R_B]$ and $E[R_A] > E[R_B]$	$A$ dominates $B$	$A$ is less risky and more rewarding, so no mean-variance investor holds $B$ instead of $A$
$SD[R_A] < SD[R_B]$ and $E[R_A] < E[R_B]$	Trade-off	More expected return costs more risk; neither dominates, and either choice can be sensible
$\Delta E[R_p] > 0$ and $\Delta SD[R_p] < 0$	Heading northwest	North is more expected return, west is less risk: higher utility for every mean-variance investor

Key Insight: Portfolio theory reads the graph without forecasting it. The points are taken as given, the prices left after everyone has traded, and the one decision left is which portfolio to build from these securities.

IIDominance or Trade-off?

One asset dominates another only if it has lower risk and higher expected return at the same time.

Asset A Asset B

IIIHeading Northwest

From any point, north means more expected return and west means less risk; northwest gains both.

Start P: SD (%) 29 P: mean (%) 10

IVIf the Gap Were a Mispricing

Buying Tansy and shorting Umber moves their prices, and a price change moves the expected return the opposite way.

Trading (long Tansy, short Umber) 0%

💡 Trading closes the spread only when it is a genuine mispricing. Long Tansy, short Umber is not riskless, and the market prices what each stock adds to a portfolio rather than its risk held alone, so on the equilibrium graph the gap can stand.

VQuiz Questions

Problem 1 · Dominance or Trade-off

Given: four stocks on an illustrative risk-return graph, each written as $(SD, \text{mean})$ in percent per year: Sorrel $(16, 6)$ , Tansy $(22, 11)$ , Umber $(39, 5)$ and Wren $(33, 9)$ . Which pair is a trade-off rather than a case of dominance?

✅ Correct! Wren pays more (

9\% > 6\%

) but is riskier (

33\% > 16\%

), so moving from Sorrel to Wren buys return with risk.

❌ Not quite. Tansy beats Wren on both axes:

22\% < 33\%

in risk and

11\% > 9\%

in mean, so Tansy dominates Wren.

❌ Not quite. Sorrel is less risky (

16\% < 39\%

) and pays more (

6\% > 5\%

), so Sorrel dominates Umber, even though its mean is only one point higher.

❌ Not quite. Wren is less risky (

33\% < 39\%

) and pays more (

9\% > 5\%

), so Wren dominates Umber.

Show solution

Compare each pair on both axes. Dominance needs the lower standard deviation and the higher mean to belong to the same stock.

Tansy vs Wren: $22\% < 33\%$ and $11\% > 9\%$ . Tansy dominates.
Sorrel vs Umber: $16\% < 39\%$ and $6\% > 5\%$ . Sorrel dominates.
Wren vs Umber: $33\% < 39\%$ and $9\% > 5\%$ . Wren dominates.
Sorrel vs Wren: $16\% < 33\%$ but $6\% < 9\%$ . The less risky stock is also the less rewarding one, so this is a trade-off.

In a trade-off either choice can be sensible: an investor who wants Wren's extra three points of expected return must accept seventeen more points of standard deviation.

Problem 2 · The Highest Return Is Not Enough

Given: on the same illustrative graph, Vetch sits at $(44, 13)$ , the highest expected return of the five stocks and also the highest standard deviation. A classmate says Vetch dominates every other stock. Is that right?

✅ Correct! To dominate a stock, Vetch would need a lower standard deviation than it, and no stock on the graph is riskier than

44\%

❌ Not quite. That reads only the vertical axis. Dominance also needs lower risk, and Vetch has the highest standard deviation on the graph.

❌ Not quite. Lower risk alone is not dominance: Tansy's mean,

11\%

, is below Vetch's

13\%

, so Tansy vs Vetch is a trade-off.

❌ Not quite. Umber's standard deviation,

39\%

, is below Vetch's

44\%

. Vetch pays more but is riskier, so this pair is a trade-off too.

Show solution

"A dominates B" needs $SD[R_A] < SD[R_B]$ and $E[R_A] > E[R_B]$ . Vetch's standard deviation, $44\%$ , is the largest on the graph, so the first condition fails against every other stock: Vetch cannot dominate anything.

Nothing dominates Vetch either, because nothing pays more than $13\%$ . Each comparison with Vetch is a trade-off, the same relation the video draws between General Motors and Motorola.

Problem 3 · Reading Directions on the Graph

Given: the illustrative graph again, as $(SD, \text{mean})$ in percent per year: Sorrel $(16, 6)$ , Tansy $(22, 11)$ , Umber $(39, 5)$ , Vetch $(44, 13)$ , Wren $(33, 9)$ .

How many of the other four stocks dominate Umber?

Switching from Sorrel to Tansy moves the investor's point which way?

✅ Correct! Sorrel, Tansy and Wren all lie to Umber's northwest, and the move from Sorrel to Tansy goes up (

6\% \to 11\%

) and right (

16\% \to 22\%

❌ Check the count. Test each stock for both conditions against Umber: standard deviation below

39\%

and mean above

5\%

❌ Check Vetch. Vetch's mean beats Umber's, but its standard deviation,

44\%

, is higher than Umber's

39\%

, so Vetch vs Umber is a trade-off.

❌ Check the direction. North is up (more expected return) and east is right (more risk). Compare Sorrel's and Tansy's numbers on each axis separately.

❌ Check the horizontal move. The mean does rise, but standard deviation also rises from

16\%

22\%

: that is a move east, not west.

Show solution

Part 1. A stock dominates Umber $(39, 5)$ when its standard deviation is below $39\%$ and its mean is above $5\%$ .

Sorrel: $16 < 39$ and $6 > 5$ . Dominates.
Tansy: $22 < 39$ and $11 > 5$ . Dominates.
Wren: $33 < 39$ and $9 > 5$ . Dominates.
Vetch: $44 > 39$ . Fails the risk test, so a trade-off.

Three stocks dominate Umber. Like McDonald's against Merck in the video, Umber looks worse than the stocks that dominate it when each is held alone.

Part 2. From Sorrel to Tansy: $\Delta E = 11\% - 6\% = +5$ points (north) and $\Delta SD = 22\% - 16\% = +6$ points (east). Northeast is a trade-off: whether it is worth it depends on how much the investor values the extra return against the extra risk.

Problem 4 · Does the Gap Have to Close?

Given: a columnist writes: "Umber sits southeast of Tansy, so the market has mispriced Umber. Go long Tansy and short Umber, and collect free money until the gap closes." Which reply matches the academic view in the video?

✅ Correct! The points are where prices settle after trading, with supply equal to demand. Held alone, Umber looks worse than Tansy, but investors hold portfolios.

❌ Not quite. Long Tansy, short Umber is not riskless; adding other securities only brings it close. Nothing forces a gap between two stocks held alone to close.

❌ Not quite. Portfolio theory makes no forecast. The graph is taken as given, and in an efficient market any predictable pattern is already in the prices.

❌ Not quite. What matters is what Umber adds to a portfolio, not how it looks alone. At equilibrium prices it is held, so nothing pushes its expected return past Tansy's.

Show solution

The video weighs two arguments. If the gap were a genuine mispricing, buying Tansy would raise its price and lower its expected return, and shorting Umber would do the reverse. That trading would close the gap, which is exactly what should happen to a mispricing.

The economist's answer is that the graph already shows the result of that trading. Each point sits at the expected return the market considers fair, with supply equal to demand. What the market prices is what a stock adds to a portfolio, not its risk held alone, so a stock that looks dominated on its own can stay where it is. The position long Tansy, short Umber is not riskless, so nothing forces the gap shut.

With prices in equilibrium, the investor's one remaining decision is which portfolio to build from these securities.

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