Data 605 Discussion 8

2 The price of a stock on a given trading day changes according to the distribution

\(\mathbf{P_x}\) = \(\left( \begin{array}{ccc} -1 & 0 & 1 & 2\\ \frac{1}{4} & \frac{1}{2} & \frac{1}{8} & \frac{1}{2} \end{array} \right)\)

Find the distribution for the change in stock price after two (independent) trading days.

Answer:

To find the distribution for the change in stock price in two days, we need to add the price change X1 + X2 = S2.

Given the distribution function m = \(\mathbf{P_x}\) above,

X1 = c(-1, 0, 1, 2)
C1 = c(0.25, 0.5, 0.125, 0.125)

X2 = X1
C2 = C1

SX1 = c(0, 0, 0, 0)
SX2 = SX1
SX3 = SX1
SX4 = SX1
SC1 = c(0, 0, 0, 0)
SC2 = SC1
SC3 = SC1
SC4 = SC1

ST = c(0, 0, 0, 0, 0, 0, 0)


SX1 = X1[1] + X2
SC1 = C1[1] * C2
SX1

## [1] -2 -1  0  1

SC1

## [1] 0.06250 0.12500 0.03125 0.03125

SX2 = X1[2] + X2
SC2 = C1[2] * C2 
SX2

## [1] -1  0  1  2

SC2

## [1] 0.1250 0.2500 0.0625 0.0625

SX3 = X1[3] + X2
SC3 = C1[3] * C2 
SX3

## [1] 0 1 2 3

SC3

## [1] 0.031250 0.062500 0.015625 0.015625

SX4 = X1[4] + X2
SC4 = C1[4] * C2 
SX4

## [1] 1 2 3 4

SC4

## [1] 0.031250 0.062500 0.015625 0.015625

Stock Price Change Distribution Range is from minimum of -2 to a maximum of 4 for the 2 days of trading. Probability distribution is as follows -

Probablity P(X=-2)

for (i in 1:4)
{
  if (SX1[i] == -2)
      ST[1] = ST[1] + SC1[i]
  if (SX2[i] == -2)
      ST[1] = ST[1] + SC2[i]
  if (SX3[i] == -2)
      ST[1] = ST[1] + SC3[i]
  if (SX4[i] == -2)
      ST[1] = ST[1] + SC4[i]  
}
ST[1]

## [1] 0.0625

Probablity P(X=-1)

for (i in 1:4)
{
  if (SX1[i] == -1)
      ST[2] = ST[2] + SC1[i]
  if (SX2[i] == -1)
      ST[2] = ST[2] + SC2[i]
  if (SX3[i] == -1)
      ST[2] = ST[2] + SC3[i]
  if (SX4[i] == -1)
      ST[2] = ST[2] + SC4[i]  
}
ST[2]

## [1] 0.25

Probablity P(X=-0)

for (i in 1:4)
{
  if (SX1[i] == 0)
      ST[3] = ST[3] + SC1[i]
  if (SX2[i] == 0)
      ST[3] = ST[3] + SC2[i]
  if (SX3[i] == 0)
      ST[3] = ST[3] + SC3[i]
  if (SX4[i] == 0)
      ST[3] = ST[3] + SC4[i]  
}
ST[3]

## [1] 0.3125

Probablity P(X=1)

for (i in 1:4)
{
  if (SX1[i] == 1)
      ST[4] = ST[4] + SC1[i]
  if (SX2[i] == 1)
      ST[4] = ST[4] + SC2[i]
  if (SX3[i] == 1)
      ST[4] = ST[4] + SC3[i]
  if (SX4[i] == 1)
      ST[4] = ST[4] + SC4[i]  
}
ST[4]

## [1] 0.1875

Probablity P(X=2)

for (i in 1:4)
{
  if (SX1[i] == 2)
      ST[5] = ST[5] + SC1[i]
  if (SX2[i] == 2)
      ST[5] = ST[5] + SC2[i]
  if (SX3[i] == 2)
      ST[5] = ST[5] + SC3[i]
  if (SX4[i] == 2)
      ST[5] = ST[5] + SC4[i]  
}
ST[5]

## [1] 0.140625

Probablity P(X=3)

for (i in 1:4)
{
  if (SX1[i] == 3)
      ST[6] = ST[6] + SC1[i]
  if (SX2[i] == 3)
      ST[6] = ST[6] + SC2[i]
  if (SX3[i] == 3)
      ST[6] = ST[6] + SC3[i]
  if (SX4[i] == 3)
      ST[6] = ST[6] + SC4[i]  
}
ST[6]

## [1] 0.03125

Probablity P(X=4)

for (i in 1:4)
{
  if (SX1[i] == 4)
      ST[7] = ST[7] + SC1[i]
  if (SX2[i] == 4)
      ST[7] = ST[7] + SC2[i]
  if (SX3[i] == 4)
      ST[7] = ST[7] + SC3[i]
  if (SX4[i] == 4)
      ST[7] = ST[7] + SC4[i]  
}
ST[7]

## [1] 0.015625