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F(x) - cumulative distribution

If you have integrated f(x) to F(x)- is your new function the cumulative function? I.e. if in 2 parts, the 2nd part will also include the area under the curve for the first part? Hope that makes sense :smile:

edit: struggling to say exactly what I mean,
say if F(x) was defined in 2 parts, say one between 1<x<3 and one 4<x<7

If you wanted to find the probability that X is less than 6 - would you only do F(6) in the 2nd function, or would you also do F(6) for the first function and add them?

(edited 5 years ago)

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Reply 1

jazz_xox_

I want to say that each 'section' of the function is only for that specific area under the curve, but then wouldn't that go against the idea of it being a cumulative function? Think I'm confusing myself here :frown:

Reply 2

TheYearNiner

Yes add them if it is less than 6. I'd explain better if it were a question. Don't forget Yates correction tomorrkw

Reply 3

jazz_xox_

Original post by TheYearNiner

Yes add them if it is less than 6. I'd explain better if it were a question. Don't forget Yates correction tomorrkw

Thank you, that's what I thought but then my friend was questioning me and I couldn't explain why.. so even though it's the 'cumulative' function, it doesn't necessarily include everything from 0, just in that section?

Usually the questions makes it clear what they're after (e.g. to add on the first section of the area under the curve to give the whole cumulative function) and I've not had trouble with it before but I'm really confusing myself now!

And thank you, I won't forget!! :smile:

Are you doing the exam too?

Reply 4

Prasiortle

Original post by jazz_xox_

If f(x) is the probability density function, then F(x) is the cumulative distribution function. However, if f(x) is defined piecewise, then you will need to add the areas to get the total (cumulative area), so to find P(X <= k) where k is less than 3, you would just do F(k) in the 1st function; to find P(X <= k) where k is more than 3, you would have to do F(3) in the 1st function to get that total area, then F(k) in the 2nd function to get that area, and then add the two areas.

Reply 5

TheYearNiner

Original post by jazz_xox_

Are you doing the exam too?

Yeah I'm doing the exam but make sure you double check with textbook because I might be wrong I can be clumsy

Reply 6

jazz_xox_

Original post by Prasiortle

Ahh that helps a lot thank you! So it's cumulative if it's just defined in one section, but if it's defined for multiple sections, you would have to combine them to find the overall cumulative function?

When you've written F(x) as multiple pieces, defined for difference x values, would you still call this the 'cumulative distribution function' or not? :smile:

Reply 7

jazz_xox_

Original post by TheYearNiner

Yeah I'm doing the exam but make sure you double check with textbook because I might be wrong I can be clumsy

Good luck

No I agree with what you're saying, I did a question earlier where you had to add on the first section of the area onto the curve to the second to show that the cumulative distribution function was equal to what they had given. I think the concept of it being 'cumulative' was confusing me as I wasn't getting why one function for F(x) didn't include the whole area.. if that makes any sense at all haha

Reply 8

Prasiortle

Original post by jazz_xox_

It's still called a cumulative distribution function.

Yes, if it's piecewise, you have to combine them. It's the same as how, e.g. if you're using a velocity-time graph that has one section between time 0 seconds and time 5 seconds, and another (different) section between time 5 seconds and time 10 seconds, and you want to find the value of t such that the displacement (i.e. area under the velocity-time graph) at time t seconds is 3 metres, and you know that this value of t is between 5 and 10, then you have to do the total area from time 0 seconds up to time 5 seconds, plus the area from time 5 seconds up to time t seconds, as this will give the total area under the graph from time 0 seconds up to time t seconds.

Reply 9

Devvo666

I'm getting confused now haha. Can someone please put an example of this type up because I can't find an example of this type in the textbook?

Reply 10

jazz_xox_

Original post by Prasiortle

Okay it's starting to make a lot more sense :smile:

If you were to define the cumulative distribution function for the 2nd section of the function only, would you still add on the area from the first part of the curve then?

Reply 11

jazz_xox_

Original post by Devvo666

I'm getting confused now haha. Can someone please put an example of this type up because I can't find an example of this type in the textbook?

I can't find an example of the question that I'm trying to answer, on all AQA papers they seem to be less complex but I was just struggling to understand the theory behind it!

http://filestore.aqa.org.uk/subjects/AQA-MS2B-QP-JUN14.PDF

7bi on this paper kind of helped me to understand it, in the way that you have to add on the first bit of the function :smile:

Reply 12

Prasiortle

Original post by jazz_xox_

Okay it's starting to make a lot more sense :smile:

If you were to define the cumulative distribution function for the 2nd section of the function only, would you still add on the area from the first part of the curve then?

The cumulative distribution function is defined piecewise: for the first section, it's just the integral from the start point up to x of the first section of the probability density function; for the second section, it's the integral over the entire domain (start point up to end point) of the first section of the probability density function, plus the integral from the second section's start point up to x of the probability density function.

Reply 13

Martha8

Original post by jazz_xox_

F(x) by definition is the integral of f(x) from -ve infinity up to x, so to find the probability that x < 6, you just need to do F(6) in the 2nd function; the height described by the first function (ie in this case the probability that x < 3) is accounted for as the '+c' when you integrate from f(x) -> F(x) for the 2nd function.

Reply 14

Devvo666

I don't understand how you get the shape of the graph for this question or the 2 parts to b. Can anyone help?

0B0CEFDA-A81D-434D-83D6-739DCCA1C590.jpg.jpeg

0B0CEFDA-A81D-434D-83D6-739DCCA1C590.jpg.jpeg

Reply 15

Prasiortle

Original post by Devvo666

I don't understand how you get the shape of the graph for this question or the 2 parts to b. Can anyone help?

You should really start your own (new) thread for this, rather than hijacking an existing thread. Secondly, your image was too small to see properly, so I had to Google around and eventually found that this was from the AQA S2 June 2017 paper. In future, pleas either make sure that the image is large enough, or state which exam board, paper, and year the question is from.

Anyway, to sketch the graph, just notice that

y = k(x+a)^{2}(x-a)^{2}

is a quartic with double roots, and thus stationary points, at

x = a

and

x = -a

(double roots due to the brackets being squared).

For (b)(i): We have

Y = 7X + 7a

, so

E(Y) = E(7X + 7a)

, which by linearity of expectation becomes

7E(X) + 7a

, and now you just have to substitute in your value of

E(X)

from part (a)(iii).

Then for (b)(ii): we have

Var(Y) = Var(7X + 7a)

, and this time we apply the rule that when

p

and

q

are constants,

Var(pX+q) = p^{2}Var(X)

, which can itself be derived from linearity of expectation. Thus we get

Var(Y) = 7^{2}Var(X) = 49Var(X)

, and again we just have to substitute in our value of

Var(X)

from part (a)(iv).

(edited 5 years ago)

Reply 16

jazz_xox_

Original post by Prasiortle

Sorry for keep asking questions.. but after you have integrated both pieces to get Fx, do you have to physically add them together to find the cumulative function for the entire domain ??

So if you’ve just integrated the second part of the function (ignoring the first part) is that just the area under that section of the curve??

Reply 17

jazz_xox_

Original post by Martha8

Ahh okay .. so you wouldn’t need to add on the first section of Fx at all?

Reply 18

Prasiortle

Original post by jazz_xox_

Ahh okay .. so you wouldn’t need to add on the first section of Fx at all?

You cannot get the cumulative distribution function as one thing, because you have to know which section of the graph your ending value falls into in order to work out the area. If your ending value falls in the first section, you will need to add the entire area of the first section to a certain area under the second section.

Reply 19

jazz_xox_

Original post by Prasiortle

Ahh okay

someone earlier in the thread says not to add on the first section as it is equivalent to the ‘+c’ ?

So my understanding now is, the whole collection of functions is the cumulative distribution function, and to find an area below a point in the section section, you also need to add on the area under the curve for the first section ? I don’t know how I’ve never noticed that this didn’t make sense to me before !