cumulative distribution function: for p(x)= (2 x)/n(n+1) where x=1,2,..,n, find cdf

[jkaks]

A discrete r.v has pdf. Let n>1
p(x)= (2 x)/n(n+1) where x=1,2,..,n

find cdf

these standard formulas can be used
∑ x= n(n+1)/2
∑ x^2=(n(n+1)(2n+1))/6

 

[Deleted member 4993]

A discrete r.v has pdf. Let n>1
p(x)= (2 x)/n(n+1) where x=1,2,..,n

find cdf

these standard formulas can be used
∑ x= n(n+1)/2
∑ x^2=(n(n+1)(2n+1))/6

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[jkaks]

I apologise for not reading the rules of the forum.
The reason i posted this question was precisely because i didn't know how to start.
I was hoping that somebody out there would set me on the right path.

thank you

 

[Deleted member 4993]

A discrete r.v has pdf. Let n>1
p(x)= (2 x)/n(n+1) where x=1,2,..,n

find cdf

these standard formulas can be used
∑ x= n(n+1)/2
∑ x^2=(n(n+1)(2n+1))/6

Do you know the definitions of:

discrete r.v.
pdf
cdf ?

 

[jkaks]

I know what they all mean
but the thing that i can't get my head round is p(x)= (2 x)/n(n+1) where x=1,2,..,n
i don't know how to work with it.

I don't know if i'm making sense
but if i had for example a pmf p(x)=x/10 where x=1,2,3
i would be able to find the cdf

in this case p(x)=2x/n(n+1) where x=1,2,...,n
where do i start

 

[pka]

I know what they all mean
but the thing that i can't get my head round is p(x)= (2 x)/n(n+1) where x=1,2,..,n
in this case p(x)=2x/n(n+1) where x=1,2,...,n CORRECT!

The CDF
\(\displaystyle C(x)=\begin{cases}0 &: x<1 \\\displaystyle\sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {p(k)} &:\left\lfloor x \right\rfloor \le x < \left\lfloor x \right\rfloor + 1\\1 &: x\ge n \end{cases}\)

 

[jkaks]

Thank you pka but i'm more interested in the process than the answer itself.
For instance how did you get (k), shouldn't it be (x)
plus i don't understand your notations on the limits

 

[Steven G]

Thank you pka but i'm more interested in the process than the answer itself.
For instance how did you get (k), shouldn't it be (x)
plus i don't understand your notations on the limits

Come on, k is just an index. You want it to be x, then let it be x.

 

[Steven G]

I know what they all mean
but the thing that i can't get my head round is p(x)= (2 x)/n(n+1) where x=1,2,..,n
i don't know how to work with it.

I don't know if i'm making sense
but if i had for example a pmf p(x)=x/10 where x=1,2,3
i would be able to find the cdf

in this case p(x)=2x/n(n+1) where x=1,2,...,n
where do i start

Note that p is a function of x and not n. So what is p(1)? What is p(2)? .... Do you see some pattern?. So what is cdf(1), cdf(2),...,cdf(n)?

 

[pka]

Thank you pka but i'm more interested in the process than the answer itself.
For instance how did you get (k), shouldn't it be (x)
plus i don't understand your notations on the limits

Reading the above reply, I am lead to think that you have no idea about any of this.
A CDF is a monotone increasing, right-continuous function that maps \(\displaystyle (-\infty,\infty)\to [0,1]\).
If the pdf is finite (as in this case), then its CDF is a finite step function.
That is exactly what I gave you. The fact that you even have to ask about the (k) in the sum, again tells me that you do not grasp the C (aCumulation) stands for in CDF.

 

[jkaks]

Reading the above reply, I am lead to think that you have no idea about any of this.
A CDF is a monotone increasing, right-continuous function that maps \(\displaystyle (-\infty,\infty)\to [0,1]\).
If the pdf is finite (as in this case), then its CDF is a finite step function.
That is exactly what I gave you. The fact that you even have to ask about the (k) in the sum, again tells me that you do not grasp the C (aCumulation) stands for in CDF.

I guess you're right,
hopefully by the end of the chapter i'll be able to understand it half as good as you.

 

[Steven G]

I guess you're right,
hopefully by the end of the chapter i'll be able to understand it half as good as you.

You need to know what CDF means (it's just a definition) when it is 1st define, not at the end of the chapter. IMO never move on until you understand theorems and definitions. After all, you can't play the game unless you know the rules (theorems and definitions).

As a student when I could not solve a problem (prove a theorem) it usually was because I did not understand a definition or a theorem.