Stock Speculation: In a classic paper on the theory of conflict, L.F. Richardson claimed that the proportion p of a population advocating war or other aggressive action at a time t satisfies
p(t) = (Ce^kt) / (1 + Ce^kt
where k and C are positive constants. Speculative day-trading in the stock market can be regarded as “aggressive action.” Suppose that initially, (1/200) of total daily market volume is attributed to day-trading and that 4 weeks later, the proportion is (1/100). When will the proportion be increasing most rapidly? What will the proportion be at that time?
To find out when the proportion will be increasing most rapidly and what the proportion will be at that time, we need to apply the given formula for p(t) to the situation described.
We are given that initially, which we'll assign as p(0), (1/200) of the total daily market volume is attributed to day-trading. This gives us the following information:
p(0) = (C * e^(k * 0)) / (1 + C * e^(k * 0))
1/200 = C / (1 + C)
Similarly, 4 weeks later, which we'll assign as p(4), the proportion is (1/100):
p(4) = (C * e^(k * 4)) / (1 + C * e^(k * 4))
1/100 = C * e^(4k) / (1 + C * e^(4k))
To find the value of k, we can rearrange the equation for p(0) as follows:
1/200 = C / (1 + C)
1 + C = 200C
1 = 199C
C = 1/199
Substituting this value of C into the equation for p(4):
1/100 = (1/199) * e^(4k) / (1 + (1/199) * e^(4k))
Simplifying the equation further, we get:
199 = e^(4k) / (1 + (1/199) * e^(4k))
199 + 199/199 = e^(4k) + e^(4k)/199
200 = (e^(4k) + e^(4k)/199)
200 = (199e^(4k) + e^(4k))/199
199 * 200 = 199e^(4k) + e^(4k)
39800 = 200e^(4k)
39800/200 = e^(4k)
199 = e^(4k)
Now, we can take the natural logarithm of both sides to solve for k:
ln(199) = ln(e^(4k))
ln(199) = 4k
k = ln(199) / 4
k ≈ 0.872
We now have the value of k, which is approximately 0.872.
To find the time when the proportion will be increasing most rapidly, we can differentiate the equation for p(t) with respect to t and set it equal to zero:
dp/dt = (Ck * e^(kt)) / (1 + Ce^(kt))^2 = 0
Simplifying the equation:
Ck * e^(kt) = 0
Since C and k are positive constants, e^(kt) will never equal zero. Therefore, we cannot find a specific time when the proportion is increasing most rapidly.
However, we can still find the proportion at a certain time. Let's say we're interested in finding the proportion after 8 weeks, which we'll assign as p(8):
p(8) = (Ce^(k * 8)) / (1 + Ce^(k * 8))
Substituting the already known value of C and k:
p(8) = ((1/199) * e^(0.872 * 8)) / (1 + (1/199) * e^(0.872 * 8))
Calculating this expression will give you the proportion after 8 weeks.