Find CDF for a PMF defined by P[X = a/n]

[Amanda_S]

I am trying to find a CDF for this PMF:

[math]\mathbb{P}\left[X=\frac{a}{n}\right] = \frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}\ for\ n=1,2,\dots[/math]
My problem is with the [imath]a \in \mathbb{R}[/imath]. This is because all examples I encountered previously are in the form [imath]\mathbb{P}\left[X=n\right][/imath]. For these types I know that the CDF becomes[imath]\mathbb{P}\left[X \leq n\right] = \sum_{i=1}^{n} \mathbb{P}\left[X=i\right][/imath] , however I am getting confused on how to go about finding the CDF for the above PMF, for it to then be used to find the quartiles.

Is there a way to derive the CDF?

Any help is greatly appreciated.

 

[BigBeachBanana]

[math]\mathbb{P}\left[X=\frac{a}{n}\right] = \frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}\ for\ n=1,2,\dots[/math]

How come there's no a in the RHS?

 

[Amanda_S]

How come there's no a in the RHS?

That's what I asked my lecturer and he said that's how he intended it to be. What is even more confusing is that later on I am supposed to find estimators for a. So I'm very lost at what to do.

I tried to find the CDF (using Mathematica) by taking the following summation [imath]\sum_{n=1}^{\frac{a}{n}} {\frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}}[/imath] as shown in the attached .pdf document, but I am still unsure as to how correct it is.

 

[BigBeachBanana]

That's what I asked my lecturer and he said that's how he intended it to be. What is even more confusing is that later on I am supposed to find estimators for a. So I'm very lost at what to do.

I tried to find the CDF (using Mathematica) by taking the following summation [imath]\sum_{n=1}^{\frac{a}{n}} {\frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}}[/imath] as shown in the attached .pdf document, but I am still unsure as to how correct it is.

If the density is not a function of a, then a has no impact. Do you have the original question?

 

[Amanda_S]

If the density is not a function of a, then a has no impact. Do you have the original question?

In the attached document is the question as well as some solutions I managed so far. I am currently attempting question 1. The working for this question so far is found in the file attached in my previous reply (which I reattached below for convenience).

 

[blamocur]

I think you can define the CDF as a function which depends on parameter 'a':
[math]F_a(t) = P(X \leq t) = P\left(X \leq \frac{a}{m}\right)[/math] where [math]m = \left\lfloor\frac{a}{t}\right\rfloor[/math][math]P\left(X \leq \frac{a}{m}\right) = \frac{6}{5} \sum_{n=m}^{\infty}\frac{6n^2}{(n+1)(n+2)(n+3)(n+4)}[/math][math]= \frac{6}{5} \frac {6m^2+9m+5}{(m+1)(m+2)(m+3)}[/math]

 

[BigBeachBanana]

Did you find the value of a?
Edit: Is the solution below yours or given to you?

 

[Amanda_S]

View attachment 32487
Did you find the value of a?
Edit: Is the solution below yours or given to you?

That solution is mine, however when I asked my lecturer he seemed okay with my answer that a can take any real value.

 

[BigBeachBanana]

That solution is mine, however when I asked my lecturer he seemed okay with my answer that a can take any real value.

To find the CDF, find the partial sum.
[math]\sum_{n=1}^{k}\frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}[/math]

 

[Amanda_S]

To find the CDF, find the partial sum.
[math]\sum_{n=1}^{k}\frac{36}{5}\frac{n^2}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}[/math]

In this case, could you please explain the reasoning behind selecting an arbitrary constant 'k' as the upper limit instead of the [imath] \frac{a}{n}[/imath], as I had done before?

Also, when it comes to finding the moments, do I take [imath] k= \frac{a}{n}[/imath], or just leave it as k?

 

[BigBeachBanana]

In this case, could you please explain the reasoning behind selecting an arbitrary constant 'k' as the upper limit instead of the [imath] \frac{a}{n}[/imath], as I had done before?

Also, when it comes to finding the moments, do I take [imath] k= \frac{a}{n}[/imath], or just leave it as k?

What troubles me is letting the upper limit to be [imath]\frac{a}{n}[/imath]. Take [imath]a=0\implies \sum_{n=1}^{0}[/imath]. That doesn't make sense to me. I need to think more about whether [imath]a \in \R[/imath] is true.