Finance-Theory · Unit 18 · Video 1 · Interactive Practice

Why Everyone's Fund Must Be the Market Portfolio

IKey Formulas

Formula	Name	What it says
$\omega_i D_j$	Investor $j$ 's holding of stock $i$	Same weight $\omega_i$ for everyone; only the dollars $D_j$ in M differ
$\sum_j \omega_i D_j = \omega_i D = V_i$	Market clearing	Total demand for stock $i$ equals its market value
$D = \sum_k V_k$	Total invested in M	Follows from $\sum_i \omega_i = 1$
$\omega_i = \dfrac{V_i}{\sum_k V_k}$	Market-portfolio weight	Each stock's share of total market capitalization

Key Insight: Two-fund separation says every investor holds T-bills plus the same fund M; supply $=$ demand then leaves exactly one candidate for M, the value-weighted market portfolio.

IITwo-Fund Separation

Every investor's risky holding is the same fund M; only the dollar amount in it differs.

Share of wealth in M (w) 70%

IIISupply Must Equal Demand

With $600bn flowing into M, only one set of weights makes the demand for every stock equal its supply.

IVThe Derivation in Symbols

Summing holdings over investors, then over stocks, pins every weight to its share of total market value.

Step 1

1 · One investor

\text{holding of stock } i \text{ by investor } j = \omega_i D_j

$D_1$ , $D_2$ , $D_3$ = $90bn, $240bn, $270bn: the dollars each investor puts into M, saved or borrowed.

VQuiz Questions

Problem 1 · One Stock Inside a Leveraged Position

Given: a toy economy with four stocks, Atlas ($300bn), Birch ($150bn), Cobalt ($90bn) and Dune ($60bn), in equilibrium. An investor with $250,000 of wealth holds 130% of it in M, borrowing the rest at the T-bill rate. Find the dollars of Cobalt she holds.

✅ Correct! Cobalt is

15\%

of M, and she has $325,000 in M, so she holds $48,750 of Cobalt.

❌ Not quite. That applies Cobalt's weight to her wealth. She holds 130% of her wealth in M, so the base is $325,000, not $250,000.

❌ Not quite. That gives Cobalt a

25\%

weight, as if M held the four stocks equally. In equilibrium M weights each stock by market value:

90/600 = 0.15

❌ Not quite. That applies Cobalt's weight to the $75,000 she borrows. The borrowed dollars go into M together with her own, so the weight applies to all $325,000.

❌ Not quite. Find Cobalt's weight in M, then multiply it by her total dollars in M.

Show solution

Step 1: Cobalt's weight in M. Total market value is $600bn, so

\omega_{\text{Cobalt}} = \frac{90}{600} = 0.15

Step 2: Dollars in M. 130% of $250,000 is $325,000, of which $250,000 is her own and $75,000 is borrowed at the T-bill rate.

Step 3: Holding of Cobalt (in dollars):

\omega_{\text{Cobalt}} D_j = 0.15 \times 325{,}000 = 48{,}750

She holds $48,750 of Cobalt. Her T-bill position is a loan of $75,000, and every other stock is held in the same proportions as in M.

Problem 2 · A Price Change Moves the Weights

Given: the same economy: Atlas $300bn, Birch $150bn, Cobalt $90bn, Dune $60bn. Atlas's share price rises $20\%$ ; no shares are issued or retired, and the other prices do not change. Find Atlas's weight in the market portfolio now.

✅ Correct! Atlas is now worth $360bn out of a total of $660bn, so

\omega_{\text{Atlas}} = 360/660 \approx 54.5\%

❌ Not quite. The weight is market value over total market value, and Atlas's market value just changed. The weight of M is not fixed; it follows prices.

❌ Close, but…

1.2 \times 50\% = 60\%

scales the numerator only. Atlas's $60bn gain also raises the total, from $600bn to $660bn.

❌ Not quite. That divides Atlas's OLD value, $300bn, by the NEW total. Both must be measured at the new price.

❌ Not quite. Recompute Atlas's market value and the total market value, then divide.

Show solution

Step 1: New market values (in billions of dollars). Atlas: $1.2 \times 300 = 360$ . The others are unchanged, so the total is

\sum_k V_k = 360 + 150 + 90 + 60 = 660

Step 2: Weight.

\omega_{\text{Atlas}} = \frac{360}{660} = \frac{6}{11} \approx 0.545

So Atlas is now about 54.5% of M. Every other weight falls; Dune's, for example, goes from $10\%$ to $60/660 \approx 9.1\%$ . An investor already holding M needs no trades: each holding changes value in step with its stock's price, so the holdings stay in proportion to market values.

Problem 3 · Clearing the Market With Three Investor Groups

Given: the same four stocks (total $600bn) are held by three groups of mean-variance investors. Group 1 has $300bn of wealth and holds 40% of it in M. Group 2 has $200bn and holds 130% of it in M, borrowing the rest. Group 3 also holds T-bills plus M. Find how many dollars group 3 must put into M for markets to clear, and how many dollars of Dune ($60bn) group 2 holds.

Group 3's dollars in M

Group 2's holding of Dune

✅ Correct! Groups 1 and 2 put $120bn and $260bn into M, so group 3 must supply the remaining $220bn of the $600bn; group 2 holds

10\%

of its $260bn in Dune, $26bn.

❌ Not quite. That subtracts the groups' wealth. Market clearing is about dollars IN M: group 1 puts 40% of $300bn there, group 2 puts 130% of $200bn.

❌ Not quite. $180bn is what group 1 lends in T-bills, not what it puts into M. Group 1's dollars in M are 40% of $300bn, $120bn.

❌ Not quite. That leaves out group 1, which also holds M: its $120bn counts toward the $600bn.

❌ Not quite. That applies Dune's weight to group 2's wealth. Group 2 has 130% of its wealth, $260bn, in M.

❌ Not quite. That uses a

25\%

weight, an equal split. Dune's weight in M is its market value over the total,

60/600 = 0.10

❌ Not quite. That applies Dune's weight to the $60bn group 2 borrows. The weight applies to all $260bn group 2 has in M.

❌ Check group 3. In equilibrium the dollars all groups put into M must add up to the total market value.

❌ Check the Dune holding. Holding = Dune's weight in M times group 2's dollars in M.

Show solution

Step 1: Dollars in M (in billions). Group 1: $0.40 \times 300 = 120$ . Group 2: $1.30 \times 200 = 260$ , of which $60$ is borrowed.

Step 2: Market clearing. Summing over stocks gave $D = \sum_k V_k$ , so the dollars in M must total the market value:

120 + 260 + D_3 = 600 \;\Rightarrow\; D_3 = 220

Group 3 puts $220bn into M.

Step 3: Group 2's Dune holding. $\omega_{\text{Dune}} = 60/600 = 0.10$ , so

\omega_{\text{Dune}} D_2 = 0.10 \times 260 = 26

Group 2 holds $26bn of Dune. Check: the three groups hold $0.10 \times (120 + 260 + 220) = 60$ , exactly Dune's $60bn.

Problem 4 · Reading a Portfolio Backwards

Given: the same equilibrium (Atlas $300bn, Birch $150bn, Cobalt $90bn, Dune $60bn). An investor's portfolio contains $13,000 of Atlas. Find how many dollars of Dune she holds, and, if her wealth is $65,000, what fraction of her wealth is in M.

Her holding of Dune

Fraction of her wealth in M

✅ Correct! Her holdings are in the ratio of market values, so Dune is

60/300 = 1/5

of her Atlas holding; $13,000 of Atlas at a

50\%

weight means $26,000 in M, which is

40\%

of $65,000.

❌ Not quite. That is 10% of her ATLAS holding. Dune's 10% weight applies to her total dollars in M, which is larger than her Atlas holding.

❌ Not quite. That scales Atlas by

90/300

, Cobalt's ratio. Dune's market value is $60bn.

❌ Not quite. That scales Atlas by

150/300

, Birch's ratio. Dune's market value is $60bn.

❌ Not quite. $13,000 out of $65,000 counts only her Atlas holding. Atlas is half of M, so her dollars in M are twice that.

❌ Not quite.

50\%

is Atlas's weight inside M, the same for every investor. Her position size is her dollars in M divided by her wealth.

❌ Close, but… that divides wealth by dollars in M. The fraction in M is dollars in M over wealth.

❌ Check the Dune holding. In equilibrium her holdings of any two stocks stand in the ratio of their market values.

❌ Check the fraction. First recover her dollars in M from the Atlas holding and Atlas's weight.

Show solution

Step 1: Ratio of holdings. Every investor holds $\omega_i D_j$ of stock $i$ , so for any two stocks the holdings are in the ratio $\omega_i / \omega_k = V_i / V_k$ :

\frac{\text{Dune holding}}{\text{Atlas holding}} = \frac{60}{300} = \frac{1}{5} \;\Rightarrow\; \frac{13{,}000}{5} = 2{,}600

She holds $2,600 of Dune.

Step 2: Dollars in M. Atlas's weight is $300/600 = 0.50$ , so

D_j = \frac{13{,}000}{0.50} = 26{,}000

Step 3: Fraction of wealth. $26{,}000 / 65{,}000 = 0.40$ , so 40% of her wealth is in M and the other $39,000 is in T-bills.

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