0

I have a sum $Y=\sum_{i=1}^{\infty}(X_i-t)u(X_i-t)$ where all $X_i's$ are i.i.d exponentially distributed random variables with parameter $\lambda$ and $t$ is a constant. I want to know how many term on average will be required in sum so that the sum is at least $L$. For this I can use following two strategies.

First Strategy:

I find the probability that the number of terms required are $k=1,2,3\cdots$ then I can use following formula $$\overline{K}=\sum_{k=1}^{\infty}k\times Pr(number \quad of \quad terms \quad required \quad are \quad k).$$

As this can be complex therefore I show another strategy below.

Second Strategy:

I find the average of individual variables (which in this case will be same because of i.i.d assumption) and then divide $L$ by the average. Hence my final answer will looklike $$\overline{K}=\frac{L}{\lambda}.$$ where $\lambda$ is average of $X_i-t$ given that $X_i>t$. I want to know if this strategy is right and if the answer $\overline{K}=\frac{L}{\lambda}$ is right.

Thanks in advance.

Frank Moses
  • 2,718
  • What's $u$? And where did "given that $X_i\gt t$" come from? – joriki Jun 02 '16 at 09:16
  • $u(x)$ is the unit step function and due to the unit step function property this given that $X_i>t$ comes – Frank Moses Jun 02 '16 at 09:23
  • 1
    The mean of $(X_i-t)u(X_i-t)$ is not the mean of $X_i-t$ given that $X_i\gt t$. In the former, values $X_i\lt t$ contribute with value $0$, whereas in the latter they don't contribute at all. – joriki Jun 02 '16 at 09:31
  • 1
    By the way, where it says "so that the sum equals $L$", you probably mean "so that the sum is at least $L$"? Since the probability of the sum being exactly $L$ is zero. – joriki Jun 02 '16 at 09:33
  • @joriki what will be the mean of $(X_i-t)u(X_i-t)$ then? I can not figure it out. can you provide some help – Frank Moses Jun 02 '16 at 09:34
  • 1
    It's $\lambda^{-1}\textsf{Pr}(X_i\ge t)$. (You used $\lambda$ twice; I'm not sure whether you meant the same thing each time; here I'm referring to $\lambda$, the parameter of the exponential distribution.) – joriki Jun 02 '16 at 09:36
  • @joriki so can I say that the average number of terms will be $\ceil{\frac{L}{\lambda^{-1}Pr(X_i>t)}}$ – Frank Moses Jun 02 '16 at 09:42
  • No, that's also clearly wrong, since it would mean that you're certain to require only $1$ term for sufficiently low $\lambda$, which isn't true. – joriki Jun 02 '16 at 09:44
  • @joriki Then how can I proceed? Using first strategy is not possible for me. Can you help me in getting the answer through any strategy – Frank Moses Jun 02 '16 at 09:49
  • @joriki Further I am confused with your comment " No, that's also clearly wrong, since it would mean that you're certain to require only 11 term for sufficiently low λλ, which isn't true". In my understanding as the $\lambda$ decreases the probability that the value of the random variable is higher than a specific number will be higher and hence, I think, we should expect that as $\lambda$ decreases the average number of terms will decrease and for sufficiently small $\lambda$ the number of terms will be $1$. – Frank Moses Jun 02 '16 at 10:00
  • No, the expected number of terms goes to $1$ for $\lambda\to0$, but it's never exactly $1$ for any non-zero $\lambda$, whereas in your expression with the ceiling it would be $1$ for sufficiently low non-zero $\lambda$. – joriki Jun 02 '16 at 10:02

1 Answers1

2

The second strategy is incorrect. You can see this by considering the case where $\lambda$ is greater than $L$. Then the expected number of terms required would be less than $1$, which is impossible.

To find the expected number of terms required, note that with probability $1-\mathrm e^{-\lambda t}$ the term is $0$ and with probability $\mathrm e^{-\lambda t}$, it is above $0$, and given that it is above $0$ it is again exponentially distributed with parameter $\lambda$. Thus the expected number $K(L)$ of terms required satisfies

$$ K(L)= 1+\left(1-\mathrm e^{-\lambda t}\right)K(L)+\mathrm e^{-\lambda t}\lambda\int_0^LK(L-\tau)\mathrm e^{-\lambda\tau}\mathrm d\tau\;, $$

or

$$ K(L)=\mathrm e^{\lambda t}+\lambda\int_0^LK(\tau)\mathrm e^{-\lambda(L-\tau)}\mathrm d\tau\;. $$

Differentiating with respect to $L$ yields

$$ K'(L)=\lambda K(L)-\lambda^2\int_0^LK(\tau)\mathrm e^{-\lambda(L-\tau)}\mathrm d\tau\;, $$

and then solving the first equation for the integral and substituting into the second yields

\begin{align} K'(L)&=\lambda K(L)-\lambda\left(K(L)-\mathrm e^{\lambda t}\right) \\ &=\lambda\mathrm e^{\lambda t}\;. \end{align}

So $K(L)=\lambda\mathrm e^{\lambda t}L+c$, and we can determine the constant $c$ by setting $L=0$ in the integral equation above, yielding $c=K(0)=\mathrm e^{\lambda t}$. Thus $K(L)=\mathrm e^{\lambda t}(\lambda L+1)$.

joriki
  • 238,052
  • Then how can I find the average number of terms so that the sum is at least $L$. Other than first strategy because I do not know the probabilities for the first strategy – Frank Moses Jun 02 '16 at 09:37
  • 1
    @FrankMoses: I added a solution. – joriki Jun 02 '16 at 10:13
  • i have tried to understand the first equation that you wrote on your answer but was unable to comprehend it. Can you explain a bit more about the reason behind every term on the right? – Frank Moses Jun 02 '16 at 14:01
  • 1
    @FrankMoses: The $1$ is for the first term you need to add no matter what the result is. Then with probability $1-\mathrm e^{-\lambda t}$ you get a value below $t$ and don't make any progress, so you need another $K(L)$ expected terms. With probability $\mathrm e^{-\lambda t}$ you get a value above $t$, and, given that it's above $t$, $X_i-t$ is exponentially distributed because of the memorylessness property of the exponential distribution. If this value is above $L$, you don't need any further terms, so no contribution from that case. If it's $\tau\in[0,L]$, you need $K(L-\tau)$ more terms. – joriki Jun 02 '16 at 14:20
  • does the first step has any particular name e.g. $xyz$ technique or $XYZ$ method. I am asking because it made the derivation lot easier and I would like to learn about it more. Thank you. – Frank Moses Jun 02 '16 at 14:47
  • 1
    @FrankMoses: It seems to sometimes be referred to as one-step analysis or first-step analysis. The general idea used is linearity of expectation. You can see another application to find the expected number of terms in a sum here. – joriki Jun 02 '16 at 15:01
  • can I use the above method to solve the problem of http://math.stackexchange.com/questions/1825239/average-number-of-terms-required-for-a-sum-of-exponential-variables-to-reach-a-s – Frank Moses Jun 14 '16 at 08:03
  • 1
    @FrankMoses: Unfortunately I don't see how to solve that problem with this method. Things get rather complicated due to the two different branches of the distribution. The solution in the present case relied on the derivative of the density being proportional to the density. You can take a look at this thread for a more involved application of the method, but I don't think the methods used there will help you with your new problem. – joriki Jun 14 '16 at 08:40
  • If the problem can not be solved by the above two methods then is there any other way of solving it? – Frank Moses Jun 14 '16 at 08:47
  • 1
    @FrankMoses: If I'd seen one, I would have said so :-) – joriki Jun 14 '16 at 08:48
  • @ Joriki : I think there is a minor mistake in your answer, see my answer on http://math.stackexchange.com/questions/1825239/average-number-of-terms-required-for-a-sum-of-exponential-variables-to-reach-a-s/1826449?noredirect=1#comment3733289_1826449 – Michael Jun 14 '16 at 22:52
  • The mistake comes when choosing the constant $c$. It should actually be $c(t)=e^{\lambda t}$, which indeed still goes to 1 when $\lambda \rightarrow 0$. – Michael Jun 14 '16 at 22:53
  • You can recover the $c(t)$ (or, more precisely, $c(\lambda, t)$) from my answer using your approach as follows: You get $K(L)=\lambda L e^{\lambda t} + c(\lambda, t)$. For fixed $\lambda$ and $t$, as $L\rightarrow 0$, the average time to exceed the threshold should be the avearge time to get the first nonzero random variable, which is $e^{\lambda t}$. So $c(\lambda, t) = e^{\lambda t}$. – Michael Jun 14 '16 at 23:05
  • @Michael: Thanks, you're right. These were actually two mistakes -- one was a typo -- where I wrote $\lambda\to0$ I meant $L\to0$. The other mistake was that I didn't take into account that there's a finite probability of taking zero steps, so $K(L)$ doesn't go to $1$ for $L\to0$ as I thought. – joriki Jun 14 '16 at 23:08
  • 2
    @Michael: I fixed the answer; I found a simple way to determine $c$. – joriki Jun 14 '16 at 23:11
  • 2
    @FrankMoses: Note that Michael found a mistake in the answer; I've corrected it accordingly. – joriki Jun 14 '16 at 23:12
  • 1
    @joriki : That fix works nicely from your previous integral equation. – Michael Jun 14 '16 at 23:22
  • @joriki yes I understand it. Thanks Joriki. – Frank Moses Jun 14 '16 at 23:33