please find 4.7.6 For random variable X of find f_X (x), E[X], and var [X]. Bern

Question

please find 4.7.6

For random variable X of find f_X (x), E[X], and var [X]. Bernoulli random variable with value p. What is the PDF f_x (x)? X is a geometric random variable with value 1/p. What is the PDF f_X (x)? When you make a phone call, the line is busy with probability 0.2 and no one answers with probability 0.3 The random variable X describes the conversation time (in minutes) of a phone call that is answered. X is an exponential random variable with E[X] = 3 minutes. Let the random variable W denote the conversation time (in seconds) of all calls (W = 0 when the line is busy or there is no answer.) (a) what is F_w (omega)? b) What is f_w(omega)? (c) w at are E[W] and Var [W]? For 80% of lectures, Professor X arrives on time and starts lecturing with delay T = 0. When Professor X is late, the starting

Explanation / Answer

Solution for 4.7.6

a) Since the conversation time cannot be negative, we know that FW (w) = 0 for w < 0.

The conversation time W is zero iff either the phone is busy, no one answers, or if the conversation time X of a completed call is zero. Let A be the event that the call is answered. Note that the event Ac implies W = 0.

For w 0,

FW (w) = P [A c ] + P [A] FW|A (w) = (1/2) + (1/2)FX (w)                                           (1)

Thus the complete CDF of W is

FW (w)       =   0                                      w < 0

           =   1/2 + (1/2)FX (w) w 0                                                                      (2)

b) By taking the derivative of FW (w), the PDF of W is

fW (w)   = (1/2)(w) + (1/2)fX (w)

=0                                              otherwise                                                        (3)

Next, we keep in mind that since X must be nonnegative, fX(x) = 0 for x < 0.

Hence, fW (w) = (1/2)(w) + (1/2)fX (w)                                                                                  (4)

c) From the PDF fW (w), calculating the moments is straightforward.

E [W]    = fW(w) dw          

             = (1/2)    fX (w) dw

             = E [X] /2                                                                                                 (5)

The second moment is

E [W2]     = w 2 fW (w) dw

= (1/2) w 2 fX (w) dw

= E[ X2 ]/2                                                                                                            (6)

The variance of W is

     Var[W] = E[ W2] (E [W])2

                 = E[ X2] /2 (E [X] /2)2                                                                                  (7)

                  = (1/2) Var[X] + (E [X])2 /4 (8)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.