Recall the set-up to derive the density function for the maximum of a Brownian motion: Start with the Brownian motion: W(0) = 0 dW satisfies: E[dW] = 0 We immediately derive that: W(T) is normal. VAR[DW] = dt E[W(T)] = 0 dW's are all normal, i.i.d. VAR[W(T)] = T M(T) = maximum of W(t) over [0, T] = MAX {W(t)|t ε [0, T]} The pdf for M(T) equals the right-hand side of the pdf for W(T) multiplied by 2. [A] Write down an integral for E[M(T)]. No need to do the integral. [B] Recall that we defined the Brownian motion's stopping time random variable, denoted Ta, as the time that the Brownian motion first crosses the value a. Here's how to sample Ta : Run the Brownian motion (do the experiment!). Wait until the first time you cross a, and write down how long you had to wait. That's the sample value of Ta for that “run” of the experiment. Recall this clever relationship for each future time t: Prob{ Ta> t } = Prob { M(t) <= a }

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Recall the set-up to derive the density function for the maximum of a Brownian motion:
Start with the Brownian motion:
W(0) = 0
dW satisfies: E[dW] = 0
We immediately derive that:
W(T) is normal. E[W(T)]
VAR[dW] = dt
= 0
VAR[W(T)] = T
M(T) = maximum of W(t) over [0, T] = MAX {W(t) | t ɛ [0, T]}
The pdf for M(T) equals the right-hand side of the pdf for W(T) multiplied by 2.
[A] Write down an integral for E[M(T)]. No need to do the integral.
[B] Recall that we defined the Brownian motion's stopping time random
variable, denoted Ta, as the time that the Brownian motion first crosses the value
a. Here's how to sample Ta : Run the Brownian motion (do the experiment!).
Wait until the first time you cross a, and write down how long you had to wait.
That's the sample value of Ta for that “run” of the experiment.
Recall this clever relationship for each future time t:
dW's are all normal, i.i.d.
Prob{ Ta > t }
= Prob {M(t) <= a }
Let H(u) denote the cumulative distribution function of Ta
(dummy variable: u.)
[A] Explain why the left-hand-side above equals 1 - H(t).
[B] Using what we know about the density function of M(t) .. can you write an
expression for H(t) as an integral?
[C] How can we get from H(u) to h(u), the pdf for Ta ?
[D] Write an expression for E[ Ta ].
Transcribed Image Text:Recall the set-up to derive the density function for the maximum of a Brownian motion: Start with the Brownian motion: W(0) = 0 dW satisfies: E[dW] = 0 We immediately derive that: W(T) is normal. E[W(T)] VAR[dW] = dt = 0 VAR[W(T)] = T M(T) = maximum of W(t) over [0, T] = MAX {W(t) | t ɛ [0, T]} The pdf for M(T) equals the right-hand side of the pdf for W(T) multiplied by 2. [A] Write down an integral for E[M(T)]. No need to do the integral. [B] Recall that we defined the Brownian motion's stopping time random variable, denoted Ta, as the time that the Brownian motion first crosses the value a. Here's how to sample Ta : Run the Brownian motion (do the experiment!). Wait until the first time you cross a, and write down how long you had to wait. That's the sample value of Ta for that “run” of the experiment. Recall this clever relationship for each future time t: dW's are all normal, i.i.d. Prob{ Ta > t } = Prob {M(t) <= a } Let H(u) denote the cumulative distribution function of Ta (dummy variable: u.) [A] Explain why the left-hand-side above equals 1 - H(t). [B] Using what we know about the density function of M(t) .. can you write an expression for H(t) as an integral? [C] How can we get from H(u) to h(u), the pdf for Ta ? [D] Write an expression for E[ Ta ].
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