Question

. Define the stochastic process Xt=2A+3Bt where PA=2=PA=-2=PB=2=PB=-2=0.5. Find PXt≥0 | t  

Answer 1

Given

Stochastic process is defined as:

where,

Then the joint probability table is:

Probabilities	A
A = -2	A = 2
B	B = -2	0.25	0.25
B =2	0.25	0.25

To determine

Solution

We have,

Four cases arise here for the values of t.

Case 1: 0 <= t < 2/3.

In this case, if A = -2, then,

(A = -2, Maximum value of B is 2 and of t is 2/3)

  

  

If A = 2, B = -2, then,

(A = 2, B = -2, t is non-negative and < 2/3)

  

  

If A = B = 2, then

(Both terms of are positive)

Hence, for 0 < t < 2/3,

  

Case 2: -2/3 < t < 0

If A = -2, B = 2, then clearly,

(Both terms of are negative)

If A = 2, B = -2, then clearly,

(Both terms of are positive)

If A = -2, B = -2, then,

(A = -2, Maximum value of B is 2, t < 2/3)

  

  

If A = 2, B = 2, then,

(A = 2, B = 2, t > -2/3 and negative)

  

  

Since, is positive in 2 of the four events in this case, for -2/3 < t < 0,

Case 3: t > 2/3

If A = -2, B = -2, then clearly,

(Both terms of are negative)

If A = 2, B = 2, then clearly,

(Both terms of are positive)

If A = -2, B = 2, then,

   (t >= 2/3)   

  

  

If A = 2, B = -2, then,

(t >= 2/3)

  

  

Since, is non-negative in 2 of the four events in this case, for t >= 2/3,

Case 4: t <= -2/3

If A = -2, B = 2, then clearly,

(Both terms of are negative)

If A = 2, B = -2, then clearly,

(Both terms of are positive)

If A = -2, B = -2, then,

   (t <= -2/3)   

  

  

If A = 2, B = 2, then,

(t <= -2/3)

  

  

Since, is non-negative in 2 of the four events in this case, for t <= -2/3,

In each of the four cases above for the values of t, we have,

Hence, we can conclude that

for