Relation between exponentially distributed random variables and Poisson(1)

  • #1
Hi,

Suppose [itex]X_1, X_2,\cdots[/itex] be an independent and identically distributed sequence of exponentially distributed random variables with parameter 1.

Now Let [itex]N_n:=\#\{1\leq k\leq n:X_k\geq \log(n)\}[/itex]

I was told that [itex]N_n\xrightarrow{\mathcal{D}}Y[/itex] where [itex]Y\sim [/itex]Poisson(1).

Could anyone give me some directions for this problem? I am totally out of mind.

Thanks.
 

Answers and Replies

  • #2
22
0
For fixed n, the condition
[tex]X_k = log ( n )[/tex]
gives you a probability for each of the exponential random variables to be included in the count. So, what distribution does that mean N has in terms of n, k and that probability?

The limit of that distribution as n goes to infinity should then be Poisson.
 
  • #3
22,089
3,286
Hi rukawakaede! :smile:

Like sfs01 suggested, you first need to calculate the distribution of Nn. That is, you'll need to know probabilities like [itex]P\{N_n=k\}[/itex]. Let's get you started on that calculation: (assume that [itex]0\leq k\leq n[/itex])

[tex]P\{N_n=k\}=P\left(\bigcup_{A\subseteq \{1,...,n\},~|A|=k}\{X_i\geq \log(n),i\in A; X_i<\log(n), i\notin A\}\right)[/tex]

Now try to calculate that using that the Xi are iid.
 
  • #4
525
5
Hi,

Suppose [itex]X_1, X_2,\cdots[/itex] be an independent and identically distributed sequence of exponentially distributed random variables with parameter 1.

Now Let [itex]N_n:=\#\{1\leq k\leq n:X_k\geq \log(n)\}[/itex]

I was told that [itex]N_n\xrightarrow{\mathcal{D}}Y[/itex] where [itex]Y\sim [/itex]Poisson(1).

Could anyone give me some directions for this problem? I am totally out of mind.

Thanks.
If you write it as [itex]N_n=\sum_{k=1}^nI(X_k \geq \log(n))[/itex] it should be easy to show that [itex]N_n[/itex] is binomial with parameter 1/n. To show convergence I'd use characteristic functions.
 
  • #5
If you write it as [itex]N_n=\sum_{k=1}^nI(X_k \geq \log(n))[/itex] it should be easy to show that [itex]N_n[/itex] is binomial with parameter 1/n. To show convergence I'd use characteristic functions.
May I know what your [itex]I[/itex] means?
 
  • #6
Hi rukawakaede! :smile:

Like sfs01 suggested, you first need to calculate the distribution of Nn. That is, you'll need to know probabilities like [itex]P\{N_n=k\}[/itex]. Let's get you started on that calculation: (assume that [itex]0\leq k\leq n[/itex])

[tex]P\{N_n=k\}=P\left(\bigcup_{A\subseteq \{1,...,n\},~|A|=k}\{X_i\geq \log(n),i\in A; X_i<\log(n), i\notin A\}\right)[/tex]

Now try to calculate that using that the Xi are iid.
Could you please tell me more about the [itex]N_n[/itex]?
 
  • #7
22,089
3,286
Uh, well, I think bpet's method is a bit easier than what I had in mind. So I suggest following him.

Anyway, the I there means a characteristic function, thus

[tex]N_n=\sum_{k=1}^n{I_{X_k^{-1}[\log(n),+\infty[}}[/tex]

where

[tex]I_A(\omega)=\left\{\begin{array}{c}0~\text{if}~\omega\notin A\\ 1~\text{if}~\omega\in A\\ \end{array}\right.[/tex]
 
  • #8
Uh, well, I think bpet's method is a bit easier than what I had in mind. So I suggest following him.

Anyway, the I there means a characteristic function, thus

[tex]N_n=\sum_{k=1}^n{I_{X_k^{-1}[\log(n),+\infty[}}[/tex]

where

[tex]I_A(\omega)=\left\{\begin{array}{c}0~\text{if}~\omega\notin A\\ 1~\text{if}~\omega\in A\\ \end{array}\right.[/tex]
If in that case, isn't the [itex]I[/itex] an indicator function?

I still can't see why [itex]N_n[/itex] is a binomial distribution.
 
  • #9
22,089
3,286
If in that case, isn't the [itex]I[/itex] an indicator function?
Yes.

I still can't see why [itex]N_n[/itex] is a binomial distribution.
[/QUOTE]

Because every indicator function IA has a Bernouilli distribution. Indeed, it becomes 0 with chance P(Ac) and it becomes 1 with chance P(A). So it is Bernouilli.

Now, the sum of Bernouilli functions is a binomial function, i.e. Bern(p)+...+Bern(p)=Bin(n,p).
Also note that Bern(p)=Bin(1,p).
All you have to do now is figure out what p is here...
 
  • #10
Yes.


Because every indicator function IA has a Bernouilli distribution. Indeed, it becomes 0 with chance P(Ac) and it becomes 1 with chance P(A). So it is Bernouilli.

Now, the sum of Bernouilli functions is a binomial function, i.e. Bern(p)+...+Bern(p)=Bin(n,p).
Also note that Bern(p)=Bin(1,p).
All you have to do now is figure out what p is here...
is the p=(1/2)n since each indicator function can be 1 or 0?
 
  • #11
22,089
3,286
No, the p of IA is P(A), since IA has a chance of P(A) of becoming 1.
Take for example throwing a weighted coin such that the chance on head is 1/3. Then it is Bernouilli(1/3) distributed. But by your reasoning, it would be Bernouilli(1/2) since the coin has two sides. It's not always so that the probability is simple 1/(number of outcomes)
 
  • #12
No, the p of IA is P(A), since IA has a chance of P(A) of becoming 1.
Take for example throwing a weighted coin such that the chance on head is 1/3. Then it is Bernouilli(1/3) distributed. But by your reasoning, it would be Bernouilli(1/2) since the coin has two sides. It's not always so that the probability is simple 1/(number of outcomes)
Hi I guess I misunderstood the meaning for p. Could you please explain again what actually p is?
 
Last edited:
  • #13
22,089
3,286
Well, a Bernouilli(p) variable is a random variable X such that X takes on two values 0 and 1, and the chance that X takes on 1 is p, and the chance that X takes on 0 is 1-p.
 

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