Baby Quantitative Finance

In this post, we will explore some basic notions in quantitative finance.

More specifically, buying and selling stocks.

Suppose 1 unit of a stock KMATH costs $1 at time t = 0. Assume negligible trading fees.

Problem 1. At time t = 0, you buy 200 units of KMATH. What is the value of your position?

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Solution. The value of $200$ units of KMATH is

$200 \times \$1 = \$200.$

Problem 2. Suppose at time t = 1, the price per unit of KMATH increased by 10%. What is the value of your position at t = 1?

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Solution. The value of the position has increased by $10\%$ , that is,

$10\% \times \$200 = 0.1 \times \$200 = \$20.$

Therefore, the position has a new value of

$\$200 + \$20 = \$220.$

Alternate Solution. If the initial position has a value of $\$P$ and the price increased by a percentage of $r_1$ , then the position increases by the value

$r_1 \times \$P = \$(P \times r_1).$

Therefore, the position would have a new value of

$\begin{aligned} \$P + \$(P \times r_1) &= \$P + r_1 \times \$P \\ &= (1 + r_1) \times \$P \\ &= \$((1 + r_1) \times P).\end{aligned}$

In particular, setting $P = 200$ and

$\displaystyle r_1 = 10\% = \frac{ 10 }{ 100 } = 0.1$

in Problem 2 yields a new value of

$\$((1 + 0.1) \times 200) = \$220.$

Problem 3. Suppose at time t = 2, the price per unit of KMATH decreased by 10%. What is the overall change in your position from t = 0 to t = 2? How about its overall percentage change? Is your position in a profit or a loss?

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Solution. We will use the calculation in Remark 1. Let

$\displaystyle r_2 = -10\% = -\frac{ 10 }{ 100 } = -0.1$

denote the percentage decrease of the price per unit of KMATH from $t = 1$ to $t = 2$ .

By Remark 1, the new position at $t = 1$ is $\$( (1+r_1) \times 200 )$ . Therefore, the new position at $t = 2$ has a value of

$\$ ( (1 + r_2) \times ((1 + r_1) \times 200) ) = \$ ( (1 + r_1) \times (1+r_2) \times 200 ).$

Substituting $r = 0.1$ , the new position has a value of

$\begin{aligned} \$ ( (1 + r_1) \times (1+r_2) \times 200 )&= \$ ( (1 + 0.01) \times (1-0.01) \times 100 ) \\ &= \$ ( (1 - 0.1^2) \times 200 ) \\ &= \$ ( (1 - 0.01) \times 200 ) \\ &= \$ 198. \end{aligned}$

The overall percentage change is

$\displaystyle -0.01 \times 100\% = -1\%.$

Since the percentage change is negative, our position currently sits in a loss.

Problem 4. For any positive integer n, let r_n denote the percentage change in your position from t = n – 1 to t = n. Show that the overall percentage change $r$ between t = 0 and t = n is calculated by

1 + r = (1 + r₁) × (1 + r₂) × … × (1 + r_n).

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Solution. Let $P_n$ denote the value of the position at time $t = n$ . Applying the alternate solution in Problem 2 repeatedly,

$\begin{aligned} \$P_n &= (1 + r_n) \times \$P_{n-1} \\ &= (1 + r_n) \times (1 + r_{n-1}) \times P_{n-2} \\ &= \, \vdots \\ &= (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times \$P_0 \\ &= \$ ( (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times P_0 ). \end{aligned}$

On the other hand, denoting the overall percentage change by $r$ , we have

$\$P_n = \$((1+r) \times P_0)$

Equating the two sides,

$\$ ((1 + r) \times P_0) = \$ ( (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times P_0 ).$

Dividing by $P_0$ on both sides yields the desired result:

$1 + r = (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1).$

Problem 5. What is the minimum percentage increase of the price per unit of KMATH from t = 2 to t = 3 required for you to not incur loss?

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Solution. Let $r_3$ denote the required percentage increase of the price per unit of KMATH from $t = 2$ to $t = 3$ . By Problem 3, the overall percentage change $r$ is given by