Expected value

[wtrow]

In a packet data communication system, a source transmits packets containing computer data
to a receiver. Because transmission errors occasionally occur, an acknowledgment (ACK) or
a nonacknowledgement (NACK) is transmitted back on a feedback channel to the source to
indicate the status of each received packet. Data packets require a highly reliable transmission
system. Assume that a packet transmission is an independent Bernoulli trial with success
probability p.

Suppose, the resource (Eg: power) expended in transmitting the packet X
times is given by (X^2 + 5) milliwatts. Find the expected amount of resource expended
in transmitting a packet successfully.


So I solved for the PMF, being p*(1-p)^(n-1) for n=1,2,...X

Now it makes sense to me that I multiply that by the X^2+5 given, however I get some nasty math:

E(n)= SUM(n*(n^2+5)*p*(1-p)^(n-1)) for n=1,...,X

This leads me to believe I am doing something wrong. Am I on the right track or not?

 

[tkhunny]

I think we're overlooking the power to send ANK/NACK, but...

Minimum: Send 1 and receive ACK -- 1^2 + 5 = 1+5 = 6 with Probability p.

One failure: Send 1 and recieve NACK, resend and receive ACK -- 9 with probability (1-p)*p

Two failure: 2*(Send 1 and recieve NACK), resend and receive ACK -- 14 with probability p*(1-p)^2

It appears to me that you have an extra 'n', in there. You're just not used to having anything but the index to calculate the mean. Stick with n^2 + 5, not n(n^2 +5).

As for the hairy math, it really isn't that bad, you just have to have a little patience.

6p + 9p(1-p) + 14p(1-p)^2 + ... + (n^2+5)p(1-p)^(n-1) + ... = S

To make things a little less hairy:

6 + 9(1-p) + 14(1-p)^2 + ... + (n^2+5)(1-p)^(n-1) + ... = S/p = T

Remember this trick?

6(1-p) + 9(1-p)^2 + 14(1-p)^3 + ... + ((n-1)^2+5)(1-p)^(n-1) + ... = T(1-p)

Now subtract

6 + 3(1-p) + 5(1-p)^2 + ... + (2n-1)(1-p)^(n-1) + ... = T - T(1-p) = S
or
3(1-p) + 5(1-p)^2 + ... + (2n-1)(1-p)^(n-1) + ... = S - 6 = W

Doesn't look liek we did anything, does it? Well, notice how we went from (n^2 + 5) to (2n - 1). Quadratic to linear! Maybe, just maybe, if we do it again, we'll reduce it to a constant!

Again, to make things a little less hairy.

3(1-p) + 5(1-p)^2 + ... + (2n-1)(1-p)^(n-1) + ... = W
3(1-p)^2 + 5(1-p)^3 + ... + (2(n-1)-1)(1-p)^(n-1) + ... = W(1-p)

Subtract

3(1-p) + 2(1-p)^2 + 2(1-p)^3 + ... + 2(1-p)^(n-1) + ... = W - W(1-p) = p(S-6)

Rewrite it a little:

[2(1-p)^2][1 + (1-p) + (1-p)^2 + ... + (1-p)^(n-3) + ...] = p(S-6) - 3(1-p)

Do you remember the sum of an infinite geometic series?

[2(1-p)^2][1/[1-(1-p)]] = p(S-6) - 3(1-p)

Simplify a bit.

[2(1-p)^2][1/p] = p(S-6) - 3(1-p)

Solving for \(\displaystyle S = \frac{5\cdot p^{2} - p + 2}{p^{2}}\)

That's all it takes for the Mean of the distribution!

Note: Don't forget to check my arithmetic!

 

[wtrow]

That makes perfect sense, thank you!