Calculating the mean of a sum of independent variables, with the same distribution, when the number of terms varies?

[MathNugget]

Suppose you have a set of max 2 independent variables, [imath]A_1 \: and \: A_2[/imath]~[imath]Exp(\lambda_i)[/imath], and the probability to go from [imath]A_1[/imath] to [imath]A_2[/imath] is [imath]b\in[0, 1][/imath].
Say A is the sum of [imath]A_1 \: and \: A_2[/imath] ([imath]A_2[/imath] can be considered 0 in the situation we don't go from [imath]A_1[/imath] to [imath]A_2[/imath]).
How would I calculate the mean of A? E(A)

It looks to me like it's a discrete function, so E(A) would be [imath]\sum outcome[/imath] x P(outcome) , for all possible outcomes.
So, it would be
E(A)=(1-b)[imath]A_1[/imath]+b([imath]A_1[/imath] + [imath]A_2[/imath]).
How am I supposed to get an actual value out of it?

 

[blamocur]

and the probability to go from A1A_1A1 to A2A_2A2 is b∈[0,1]b\in[0, 1]b∈[0,1].

Does this mean that you pick [imath]A_2[/imath] with probability [imath]b[/imath] or [imath]A_1[/imath] with probability [imath]1-b[/imath] ?

 

[MathNugget]

Does this mean that you pick [imath]A_2[/imath] with probability [imath]b[/imath] or [imath]A_1[/imath] with probability [imath]1-b[/imath] ?

I was trying to say [imath]A_1[/imath] happens certainly, but [imath]A_2[/imath] might happen (with probability b). I think it would be simpler to say it's a Bernoulli (coin flip) whether [imath]A_2[/imath] happens, while [imath]A_1[/imath] is a given to occur.
I guess I should have quoted the problem more clearly (for context):
[imath]A_1[/imath] is the time it takes event 1 to happen (it certainly happens).
[imath]A_2[/imath] is the time it takes event 2 to happen (if the coinflip of probability b gives it a go).

So the 2 possible outcomes are: just first event happens (and A = [imath]A_1[/imath]) or both happen (and A is the sum of [imath]A_1[/imath] and [imath]A_2[/imath] )

 

[BigBeachBanana]

If I understand the problem statement correctly...

[imath]E(A) = E(A_1) + E(A_2) = A_1+\dfrac{1}{2\lambda}[/imath]

Details below:
[imath]E(A_1) = A_1 \text{ as this happens certainly}[/imath]

Let [imath]X[/imath] the r.v represents the outcome of the coin flip. Assume [imath]X[/imath] is uniformly distributed over [imath][0,1][/imath], the pdf [imath]f(x) = 1; 0 \le x \le 1[/imath]

[imath]\text{By conditional expectation, } E(A_2) = E[E(A_2|X=x)][/imath] and [imath]E(A_2|X=x) = x\cdot \dfrac{1}{\lambda}[/imath]
[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

Thus,
[math]E(A_2) = \int_0^1 E(A_2|X=x) \times f(x) \, dx =\int_0^1 x\cdot \dfrac{1}{\lambda}\, dx = \dfrac{1}{2\lambda}[/math]

 

[blamocur]

I was trying to say [imath]A_1[/imath] happens certainly, but [imath]A_2[/imath] might happen (with probability b). I think it would be simpler to say it's a Bernoulli (coin flip) whether [imath]A_2[/imath] happens, while [imath]A_1[/imath] is a given to occur.
I guess I should have quoted the problem more clearly (for context):
[imath]A_1[/imath] is the time it takes event 1 to happen (it certainly happens).
[imath]A_2[/imath] is the time it takes event 2 to happen (if the coinflip of probability b gives it a go).

So the 2 possible outcomes are: just first event happens (and A = [imath]A_1[/imath]) or both happen (and A is the sum of [imath]A_1[/imath] and [imath]A_2[/imath] )

So you pick [imath]A_1[/imath] with probability [imath]1-b[/imath] and [imath]A_1+A_2[/imath] with probability [imath]b[/imath].

Let's simplify slightly: two variables, not necessarily independent, [imath]B_1[/imath] and [imath]B_2[/imath] with CDFs [imath]F_1, F_2[/imath] are picked with probabilities [imath]1-b[/imath] and [imath]b[/imath], where the picking happens independently of [imath]B_1[/imath] and [imath]B_2[/imath]. Let's call the resulting variable [imath]B[/imath]. To compute CDF of B we ask what is the probability that [imath]B \leq a[/imath] ?

We can consider 4 events:

• [imath]v_1[/imath] : [imath]B_1[/imath] is picked with probability [imath]1-b[/imath],
• [imath]v_2[/imath] : [imath]B_2[/imath] is picked with probability [imath]b[/imath],
• [imath]v_3[/imath] : [imath]B_1 \leq a[/imath] with probability [imath]F_1(a)[/imath]
• [imath]v_4[/imath] : [imath]B_2 \leq a[/imath] with probability [imath]F_2(a)[/imath]

We are interested in probability of [imath](v_1 \cap v_3) \cup (v_2 \cap v_4)[/imath]. But since [imath]v_1[/imath] and [imath]v_2[/imath] do not intersect we can claim that:
[math]P((v_1 \cap v_3) \cup (v_2 \cap v_4)) = P(v_1 \cap v_3) + P(v_2 \cap v_4)[/math]and since [imath]v_1,v_2[/imath] are independent of [imath]v_3,v_4[/imath] :
[math]P((v_1 \cap v_3) = P(v_1)P(v_3) = (1-b) F_1(a)[/math][math]P(v_2 \cap v_4) = P(v_2)P(v_3) = b F_2(a)[/math]which means that the CDF of [imath]B[/imath] is [imath](1-b)F_1 + bF_2[/imath].

Is this helpful?

 

[MathNugget]

sort of. I think I understand it (and I also think the initial problem was much simpler than I initially understood, maybe).
Seems like E(A)= [imath](1-b) E(A_1) + b E(A_2)[/imath] .
And then, since [imath]A_i[/imath] are [imath]Exp(\lambda_i)[/imath], I have take the mean from there, put it in, and seems that's it...
The exercise was asking me to compare the theoretical result to the empirical one, and I am confused as to how that's possible. The empirical result is a number, the theoretical one is some formula.
Or maybe noting the empirical one with [imath]\sigma[/imath], I have to compute P([imath]E(A)<\sigma[/imath]) and P([imath]E(A)>\sigma[/imath]), and show they're close to 0?

 

[blamocur]

The exercise was asking me to compare the theoretical result to the empirical one, and I am confused as to how that's possible. The empirical result is a number, the theoretical one is some formula.

Pick some values of [imath]b,\lambda_1, \lambda_2[/imath], get an empirical number (usually by using random number generator) for them, then plug them into your formula and compare the results. The larger your random samples used for empirical values the small is the difference between empirical and theoretical results.

 

[MathNugget]

I see. Thank you.