Question

5. Let {X(t)} be standard Brownian motion. Define {Z(t)} identical to standard Brownian motion until the first time it hits a given value a > 0 then from there on it remains equal to that value a. Compute P(Z(t) ? x) for t > 0 and x > 0. The value of this probability when t = x = 1 and a = 1/2 is closest to: (A) 0.51 (B) 0.25 (C) 0.82 (D) 1.0 (E) 0.34

          5. Let {X(t)} be standard Brownian motion. Define {Z(t)} identical to standard Brownian motion until the first time it hits a given value a > 0 then from there on it remains equal to that value a. Compute P(Z(t) ? x) for t > 0 and x > 0. The value of this probability when t = x = 1 and a = 1/2 is closest to:
(A) 0.51
(B) 0.25
(C) 0.82
(D) 1.0
(E) 0.34

5. Let X(t) be standard Brownian motion. Define Z(t) identical to standard Brownian motion until the first time it hits a given value a > 0 then from there on it remains equal to that value a. Compute P(Z(t) ? x) for t > 0 and x > 0. The value of this probability when t = x = 1 and a = 1/2 is closest to:
(A) 0.51
(B) 0.25
(C) 0.82
(D) 1.0
(E) 0.34

Added by Alex C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Shaiju T

06/15/2022

Step 1

In this case, a = 1/2. So, Z(t) will be equal to X(t) until it hits 1/2, and then will remain equal to 1/2. Show more…

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5. Let {X(t)} be standard Brownian motion. Define {Z(t)} identical to standard Brownian motion until the first time it hits a given value a > 0 then from there on it remains equal to that value a. Compute P(Z(t) ≤ x) for t > 0 and x > 0. The value of this probability when t = x = 1 and a = 1/2 is closest to: (A) 0.51 (B) 0.25 (C) 0.82 (D) 1.0 (E) 0.34