Relation between exponentially distributed random variables and Poisson(1)

[rukawakaede]

Hi,

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

Now Let $N_n:=\#\{1\leq k\leq n:X_k\geq \log(n)\}$

I was told that $N_n\xrightarrow{\mathcal{D}}Y$ where $Y\sim$Poisson(1).

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

Thanks.

 

[sfs01]

For fixed n, the condition
$$X_k = log ( n )$$
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.

 

[micromass]

Hi rukawakaede!

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

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

Now try to calculate that using that the X_i are iid.

 

[bpet]

Hi,

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

Now Let $N_n:=\#\{1\leq k\leq n:X_k\geq \log(n)\}$

I was told that $N_n\xrightarrow{\mathcal{D}}Y$ where $Y\sim$Poisson(1).

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

Thanks.

If you write it as $N_n=\sum_{k=1}^nI(X_k \geq \log(n))$ it should be easy to show that $N_n$ is binomial with parameter 1/n. To show convergence I'd use characteristic functions.

 

[rukawakaede]

If you write it as $N_n=\sum_{k=1}^nI(X_k \geq \log(n))$ it should be easy to show that $N_n$ is binomial with parameter 1/n. To show convergence I'd use characteristic functions.

May I know what your $I$ means?

 

[rukawakaede]

Hi rukawakaede!

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

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

Now try to calculate that using that the X_i are iid.

Could you please tell me more about the $N_n$?

 

[micromass]

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

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

where

$$I_A(\omega)=\left\{\begin{array}{c}0~\text{if}~\omega\notin A\\ 1~\text{if}~\omega\in A\\ \end{array}\right.$$

 

[rukawakaede]

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

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

where

$$I_A(\omega)=\left\{\begin{array}{c}0~\text{if}~\omega\notin A\\ 1~\text{if}~\omega\in A\\ \end{array}\right.$$

If in that case, isn't the $I$ an indicator function?

I still can't see why $N_n$ is a binomial distribution.

 

[micromass]

If in that case, isn't the $I$ an indicator function?

Yes.

I still can't see why $N_n$ is a binomial distribution.

[/QUOTE]

Because every indicator function I_A has a Bernouilli distribution. Indeed, it becomes 0 with chance P(A^c) 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...

 

[rukawakaede]

Yes.Because every indicator function I_A has a Bernouilli distribution. Indeed, it becomes 0 with chance P(A^c) 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?

 

[micromass]

No, the p of I_A is P(A), since I_A 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)

 

[rukawakaede]

No, the p of I_A is P(A), since I_A 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?

 

[micromass]

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.