# I Finding CDF given boundary conditions (simple stats and calc)

#### bodycare

Summary
Simple statistics and calc problem. Given two boundary conditions of a cdf how does one find the interval of the cdf.
I'm not quite sure if my problem is considered a calculus problem or a statistics problem, but I believe it to be a statistics related problem. Below is a screenshot of what I'm dealing with. For a) I expressed f(t) in terms of parameters p and u, and I got: $$f(t)=\frac{-u \cdot a + u \cdot b + a}{(\frac{1}{p \cdot(p-1)}+\frac{1}{p}+u \cdot a - u \cdot b - a)\cdot(\frac{1}{p \cdot (p-1)}+\frac{1}{p})}$$

I am unsure how to use this to find the values for a and b. I assume (hence my post in the stats forum) that I need to use the fact that f(t) is a cdf. I was thinking that since f(t) is a cdf that f(0)=0 and f(D)=M, since 0 work would be done at time t=0 and all the work would be done after D days, but this is just a thought. I need with finding the values a and b which I believe will set me up for solving b).

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#### StoneTemplePython

Gold Member
can you explain the meaning of

D as the total number of days left to complete the task
what does that mean? By any interpretation I can think of, D is not a constant and is a function of time elapsed thus far, i.e. $t$, but that contradicts that statement that

the constant C is given by
$C = \frac{\sqrt{D(D+4)} - D}{2}$
because $C$ varies with $D$ and hence is a function of time too and thus not a constant.

- - - - -
I can make guesses on how to properly interpret the above, but that is not very productive.

#### bodycare

can you explain the meaning of

what does that mean? By any interpretation I can think of, D is not a constant and is a function of time elapsed thus far, i.e. $t$, but that contradicts that statement that

because $C$ varies with $D$ and hence is a function of time too and thus not a constant.

- - - - -
I can make guesses on how to properly interpret the above, but that is not very productive.
I actually wondered the same thing. The picture I posted is all the information that is given. Although the problem set is based of a model which can be found in this edition of PT https://physicstoday.scitation.org/doi/pdf/10.1063/PT.3.3064

#### StoneTemplePython

Gold Member
I actually wondered the same thing. The picture I posted is all the information that is given. Although the problem set is based of a model which can be found in this edition of PT https://physicstoday.scitation.org/doi/pdf/10.1063/PT.3.3064
hmmmm, what did you mean by "with slightly different meanings" which was in your post but seems to be edited out?

If the meanings are the same, then the picture in the link is actually quite helpful...

#### bodycare

hmmmm, what did you mean by "with slightly different meanings" which was in your post but seems to be edited out?

If the meanings are the same, then the picture in the link is actually quite helpful...
Not sure If I understand. I mean in PT N(t) is number of submissions received by day t whereas N(t) in my problem set is the amount of work done after t days. To me those descriptions of N(t) are quite similar, but not exactly the same.

#### StoneTemplePython

Gold Member
Not sure If I understand. I mean in PT N(t) is number of submissions received by day t whereas N(t) in my problem set is the amount of work done after t days. To me those descriptions of N(t) are quite similar, but not exactly the same.
I suppose I wouldn't dwell on it. The author seems to be using $N(t)$ as an absolutely continuous random variable, though of course this isn't true of any counting variable or counting process. The approximation used by the author may be easier to work with though.

So you have a pdf
$f(t) = \frac{t}{(D+C-t)(D+C)} = \frac{1}{D+C-t} -\frac{1}{D+C}$

and a CDF for $x \in [0, D]$
$F(x) = \alpha \int_0^x \frac{t}{(D+C-t)(D+C)} dt= \alpha \int_0^x \big(\frac{1}{D+C-t} -\frac{1}{D+C}\big)dt$
or
$F(x) = \alpha \big(\int_0^x \frac{1}{D+C-t}dt\big) - \alpha\frac{x}{D+C}$

you can directly verify that this is zero when $x = 0$
you now need to solve for $\alpha$ such that this is 1 when x is set equal to D

edit:
I seems to have misread this. The original screenshot defines f(t) to in effect be the CDF. This is extremely misleading from a notational standpoint, but it checks out.

if f(t) is infact to be the cdf (cumulative percentage of work completed) then evaluating at t=D gives
$\frac{1}{D}-\frac{1}{D+C} = 1$
which holds, e.g. here

https://www.wolframalpha.com/input/?i=1/((sqrt(d(d+4))+-+d)/2)+-+1/(d+++(sqrt(d(d+4))+-+d)/2+)

So it is a valid CDF. There are no 'boundary conditions' needed to normalize this CDF in this form. It seems that OP's issue is to do some substitutions and figure out the implied values of $u$ and $p$ from here.

cleanup edit:
in the edit section, I meant to type
$f(D) = \frac{1}{D +C - D} - \frac{1}{C+D} = \frac{1}{C } - \frac{1}{D+C}$
the wolfram link lines up with this

Last edited:
• bodycare

#### bodycare

I suppose I wouldn't dwell on it. The author seems to be using $N(t)$ as an absolutely continuous random variable, though of course this isn't true of any counting variable or counting process. The approximation used by the author may be easier to work with though.

So you have a pdf
$f(t) = \frac{t}{(D+C-t)(D+C)} = \frac{1}{D+C-t} -\frac{1}{D+C}$

and a CDF for $x \in [0, D]$
$F(x) = \alpha \int_0^x \frac{t}{(D+C-t)(D+C)} dt= \alpha \int_0^x \big(\frac{1}{D+C-t} -\frac{1}{D+C}\big)dt$
or
$F(x) = \alpha \big(\int_0^x \frac{1}{D+C-t}dt\big) - \alpha\frac{x}{D+C}$

you can directly verify that this is zero when $x = 0$
you now need to solve for $\alpha$ such that this is 1 when x is set equal to D

edit:
I seems to have misread this. The original screenshot defines f(t) to in effect be the CDF. This is extremely misleading from a notational standpoint, but it checks out.

if f(t) is infact to be the cdf (cumulative percentage of work completed) then evaluating at t=D gives
$\frac{1}{D}-\frac{1}{D+C} = 1$
which holds, e.g. here

https://www.wolframalpha.com/input/?i=1/((sqrt(d(d+4))+-+d)/2)+-+1/(d+++(sqrt(d(d+4))+-+d)/2+)

So it is a valid CDF. There are no 'boundary conditions' needed to normalize this CDF in this form. It seems that OP's issue is to do some substitutions and figure out the implied values of $u$ and $p$ from here.
I believe you made a small mistake.
How did you get $\frac{1}{D}-\frac{1}{D+C} = 1$
When I plug in t=D I get $f(D)=\frac{1}{C}-\frac{D}{D+C}$

#### StoneTemplePython

Gold Member
I believe you made a small mistake.
How did you get $\frac{1}{D}-\frac{1}{D+C} = 1$
When I plug in t=D I get $f(D)=\frac{1}{C}-\frac{D}{D+C}$
plug in: $C = \big(\sqrt{(D(D+4))} - D\big)\cdot \frac{1}{2}$ -- as defined in your original post
Then click the wolfram link in my post

#### bodycare

plug in: $C = \big(\sqrt{(D(D+4))} - D\big)\cdot \frac{1}{2}$ -- as defined in your original post
Then click the wolfram link in my post
I agree with your wolfram link since that included the fraction $\frac{1}{C} - \frac{1}{D+C}$ but what you wrote in the post was $\frac{1}{D} - \frac{1}{D+C}$

#### StoneTemplePython

Gold Member
I agree with your wolfram link since that included the fraction $\frac{1}{C} - \frac{1}{D+C}$ but what you wrote in the post was $\frac{1}{D} - \frac{1}{D+C}$
yes -- typo in my post, but not in the wolfram calculation. it happens.

#### bodycare

yes -- typo in my post, but not in the wolfram calculation. it happens.
I apologize for nitpicking. I am unsure how to find the values for a and b. When I did my substitution with p and u I got

$f(t)=\frac{-u \cdot a + u \cdot b + a}{(\frac{1}{p \cdot(p-1)}+\frac{1}{p}+u \cdot a - u \cdot b - a)\cdot(\frac{1}{p \cdot (p-1)}+\frac{1}{p})}$

but I am unable to go on from here.

"Finding CDF given boundary conditions (simple stats and calc)"

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