- #1

wildman

- 31

- 4

Could someone give me a hint for this problem? I have no idea how to attack it.

The random variables A and B are independent [tex] N(0;\sigma) [/tex] and p is the probability that the process X(t) = A - Bt crosses the t axis in the interval (0,T). Show that [tex] \pi p = arctan T [/tex]. Hint: p = P{0 <= A/B <= T}.

Well I know that independent means:

[tex] f(x,y) = f_x(x) f_y(y) [/tex]

Normal is equal to:

[tex] \frac{1}{\sqrt{2\pi {\sigma}^2}} e^{\frac{-x^2}{2 {\sigma}^2}} [/tex]

The hint says that the answer is the probability that A/B is between 0 and T.

The random variables A and B are independent [tex] N(0;\sigma) [/tex] and p is the probability that the process X(t) = A - Bt crosses the t axis in the interval (0,T). Show that [tex] \pi p = arctan T [/tex]. Hint: p = P{0 <= A/B <= T}.

Well I know that independent means:

[tex] f(x,y) = f_x(x) f_y(y) [/tex]

Normal is equal to:

[tex] \frac{1}{\sqrt{2\pi {\sigma}^2}} e^{\frac{-x^2}{2 {\sigma}^2}} [/tex]

The hint says that the answer is the probability that A/B is between 0 and T.

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