For the cdf F(x) find the pmf f(x), 25th percentile, 60th percentile

[RJLiberator]

Homework Statement

We have F(x) = ∑ (1/2)^j [Sum goes from j=1 to x]
For the cdf F(x), find the pmf f(x), the 25th percentile, and the 60th percentile.

Homework Equations

The Attempt at a Solution

I've been doing numerous of these for continuous distributions, however this one is tricking up my understanding.

So we have a CDF that is being summed here, but a CDF function can only go from 0 to 1.
So, would that mean that x mus equal 2, since 1/2 + 1/2 = 1?

Thus, the graph of F(x) just looks like 0 until it reaches 1, then it jumps up to 1/2 until it reaches x = 2 and stabilizes at 1.

Thus, the PDF is then 1/2 at x = 1 and 2, and 0 elsewhere.

Thus, the 25th percentile is 0.
The 60th percentile is x = 1 as it attains a value of 0.5 which is less than 0.6.

Do I have my understanding correct?

 

[StoneTemplePython]

Homework Statement

We have F(x) = ∑ (1/2)^j [Sum goes from j=1 to x]
For the cdf F(x), find the pmf f(x), the 25th percentile, and the 60th percentile.

Homework Equations

The Attempt at a Solution

I've been doing numerous of these for continuous distributions, however this one is tricking up my understanding.

So we have a CDF that is being summed here

No. Your CDF is summing your PMF.

Thus, the PDF is then 1/2 at x = 1 and 2, and 0 elsewhere.

Thus, the 25th percentile is 0.
The 60th percentile is x = 1 as it attains a value of 0.5 which is less than 0.6.

Do I have my understanding correct?

Also no. You have a PMF here not a PDF (which your problem statement even says) and that is not what it looks like.

What does the PMF look like? (Hint: it looks like tossing a coin until...)

 

[RJLiberator]

Sorry, my teacher calls PMF and PDF the same thing while my book considers them differently, that is why I mix up the pdf/pmf language.

So, are you saying the PMF looks like tossing a coin until it records a value of 1? Essentially it gets 0 or 1 and that means the average would be 1/2. But, once it reaches the value of 1, then the CDF maximum value is obtained as the CDF can't go beyond value of 1?

 

[StoneTemplePython]

There's basically two ways to do this. (I can think of more than 2 but let's go with these 2.)

1.) A physical interpretation of what's going on. What kind of experiment would give you the underlying PMF? It has something to do with coin tossing until getting a heads (1). Draw a sketch of this process. After sketching out the underlying graph (i.e. bar chart of pmf probabilities) you should not think there is a mean of $\frac{1}{2}$. Think very carefully about the physical process here.

2.) Perhaps more math-y: working directly and symbolically with the CDF. You can (a) get a closed form for this CDF if you recognize the underlying finite series. And irrespective of that, you can do (b): look at the differences $F_X(x+1) - F_X(x)$ to figure out the implied PMF contribution is at each 'up tick'. (This is, loosely, discrete differentiation since each step size is one so there would be a one in the denominator in the difference quotient if you were so inclined. But you can ignore this and just look at the graph of the CDF and get the same idea.)
- - - -
in either case you need to sketch this out. Drawing a picture, perhaps multiple pictures, is vital.

 

[Ray Vickson]

Homework Statement

We have F(x) = ∑ (1/2)^j [Sum goes from j=1 to x]
For the cdf F(x), find the pmf f(x), the 25th percentile, and the 60th percentile.

Homework Equations

The Attempt at a Solution

I've been doing numerous of these for continuous distributions, however this one is tricking up my understanding.

So we have a CDF that is being summed here, but a CDF function can only go from 0 to 1.
So, would that mean that x mus equal 2, since 1/2 + 1/2 = 1?

Thus, the graph of F(x) just looks like 0 until it reaches 1, then it jumps up to 1/2 until it reaches x = 2 and stabilizes at 1.

Thus, the PDF is then 1/2 at x = 1 and 2, and 0 elsewhere.

Thus, the 25th percentile is 0.
The 60th percentile is x = 1 as it attains a value of 0.5 which is less than 0.6.

Do I have my understanding correct?

For a discrete random variable taking values in $\{1,2,3, \ldots\}$ the connection between the pmf $p(y)$ and the CDF $F(x)$ is that
$$F(x) = \sum_{y=1}^x p(y).$$
So, getting your pmf $p(\cdot)$ ought to be a snap.

As for finding the percentiles: just evaluate the summation explicitly; it is an elementary result because $p!1), p(2), p(3), \ldots$ is a geometric sequence. That leaves you with two straightforward equation-solving tasks.