Question

Let A and B be two discrete random variables with joint PMF....

Answer 1

Let A and B be two discrete random variables with joint PMF \[P _{A,B} (n,m).\\ What\ is \ \lim_{n \rightarrow \infty} P _{A,B} (n,0)\]

Answer 2

I'm thinking Py(0)?

Answer 3

A good hard look at two Poisson Distributions could be a useful exploration. Very convenient Joint Distribution.

Answer 4

I understand, but this is a case of discrete RVs. I'd like my argument to be based on discrete.

Answer 5

@agent0smith

Answer 6

@Hero

Answer 7

?? Poisson is discrete. That's why I selected it.

Answer 8

@tkhunny True, sorry. So I'm thinking it's 0?

Answer 9

That's what I was thinking.

Answer 10

This is my reasoning: The probability of one part of A will always be less than that of the total and since A is constrained to one value of B, stretching A to infinity is just like a 1 dimensional random variable whose limit at inf is 0. So the final result is 0.

Answer 11

If that's wrong, I'm not getting it. Good work.

Answer 12

It's all you. Can I ask another more intricate question?

Answer 13

The probability of the amount of time taken for a secretary to process a memo independent of others is modeled as an exponential random variable with PDF
\[\ f_{T}(t) = \frac{ 1 }{ 2 }e ^{-\frac{ t }{ 2 }} \]

Answer 14

Also, the probability of the number of memos that the secretary is assigned daily is modeled as a Poisson RV with PMF
\[P_{N}(k) = \frac{ L^k}{ k! }e^{-L} \mbox{ for all integers }k\ge 0\]

Answer 15

What is the total expected amount of time the secretary spends on memos per day?

Answer 16

@ganeshie8 @tkhunny

Answer 17

?? My first impression is that the answer is outside the calculation environment. Does the secretary do anything else. 7.5 hour day suggests 7.5 hours.

The exponential distributions are independent. What's the mean of that distribution?

Answer 18

Exactly my concern. I pointed out that since once does not know the interval between the memos and whether the secretary can perform multiple jobs at the same time (so the total time will simply be the max) and other concerns.
But I'm assuming we're to just add the individual times, so say memo 1 takes t1 and memo 2 takes t2; even if there's an interval between, just add t1 t2.
If you have another interpretation, feel free to share.
The mean if an exp distribution is 1/Lambda, or 1/1/2 = 2 in this case.

Answer 19

Not a bad idea, but it takes a little more care.
E(Hrs|0 Memo) = 0
E(Hrs|1 Memo) = 2
E(Hrs|2 Memo) = 4
E(Hrs|3 Memo) = 6
E(Hrs|4 Memo) = 8 -- This may need to be 7.5, depending on the work day.
E(Hrs|5 Memo) -- Again, only 7.5???

This is a truncated Exponential distribution, so the mean is NOT 2.

Answer 20

This may be more sophistication than the original problem intended, but I think the mean of the exponential distribution is only about 1.82 hours.

Answer 21

Wait, you're factoring in the real time? Let's assume the day spans an infinite period. I think it was only used for illlustrative purposes.

Answer 22

That's what I was talking about as far as "more sophistication." If we assume infinite time, we just have a sum of exponentials. How many?

Answer 23

How many exponentials? The number of memos is given as a separate probability based on Poisson.

Answer 24

...and its mean is??

Answer 25

L.

Answer 26

@Zarkon !!!

Answer 27

There you have it. And we need the mean of the sum of L exponentials.

Answer 28

@tkhunny I thought, just use E(NT) = E(T)E(N) = 2L, but that seems faulty.

Answer 29

Here's why:

Answer 30

@Zarkon Feel free to contribute too. You were really helpful last time.