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(math246)[2004](f)test2~PPSpider^sol_10494.pdf
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MATH 246 Probability and Random Processes
Solution to Test Two
Fall 2004 Course Instructor: Prof. Y. K. Kwok
1. Let X be the time (in days) between consecutive accidents, and X is exponential with parameter , satisfying = 2. The distribution function of X is
.2t
FX (t)=1 . e ,t> 0.
.4
Hence, P [X> 2] = 1 . P [X 2] = 1 . FX (2) = e.
2.
The ratio P [N = k]/P [N = k . 1] = /k, which decreases with increasing k. For < 1, /k < 1 for k 1 so that P [N = k] attains its maximum value at k = 0. Suppose 1 and is not an integer, then the ratio is greater than 1 at k = .oor() and becomes less than 1 at .oor () + 1. When happens to be an integer, P [N = ]/P [N = . 1] = 1. The maximum value of P [X = k] occurs at both k . 1 and k.
3.
Recall that P [X = k]= pqk.1 and P [X>j]= qj
k+j.1
P [X = k + j,X >j] pq
k.1
P [X = k + j|X>j]= == pq = P [X = k].
P [X>j] qj
If a success has not occurred in the earlier j trials, then the probability of having to perform exactly k more trials to get a success is the same as the probability of initially having to perform exactly k trials to get a success. This is related to the memoryless properties of the geometric random variable.
4. When .1 <x< 1, that is, 0 <y< 1,y = x2 has two roots, namely, x1 = . y and x2 = y. Therefore,
1 .2 2 . 2
fY (y)= (1 + y)+ (1 . y)= .
2y 99 9y
When 1 x< 2, that is, 1 y< 4,y = x2 is strictly increasing. We have
.
dx .
2 11 .1 .
fY (y)= fX (x)=(x +1)= 1+ .
dy 92x 9 y
In summary,
2
. 0 <y< 1
9 y
.
.
fY (y)= 1 .1+ 1. 1 y< 4 .
9 y
.
. 0 otherwise
1