# Continuous Random Variables HelpWatch

Thread starter 2 weeks ago
#1
f(x) = kx(x-50)^2(100-x) for 0 < x < 100 (or equal to)
0 otherwise.

i) Show that k = 1.2 x 10^-8
ii) Sketch the graph of f(x)

I have done both of these questions.

iii) Describe the shape of the graph and give an explaination of how such a graph might occur, in terms of the examination and the candidates
iv) Is it permissible to model a mark, which is a discrete random variable going up in steps of 1, by a continuous random variable like X, as defined in this question?

These have got me really confused. Any help?
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2 weeks ago
#2
(Original post by TAEuler)
f(x) = kx(x-50)^2(100-x) for 0 < x < 100 (or equal to)
0 otherwise.

i) Show that k = 1.2 x 10^-8
ii) Sketch the graph of f(x)

I have done both of these questions.

iii) Describe the shape of the graph and give an explaination of how such a graph might occur, in terms of the examination and the candidates
iv) Is it permissible to model a mark, which is a discrete random variable going up in steps of 1, by a continuous random variable like X, as defined in this question?

These have got me really confused. Any help?
As far (iii) is concerned, I think it's fairly obvious what type of equation that is. I mean... if you expand the whole thing, what's the highest power you're going to have? Will it have a positive or negative coefficient? So how would it look like?? You can snipe out the roots of it too since it's in a factorised form already, so you should have a good idea of how it looks like.

For its second part of (iii), well clearly it is referring to some context which you haven't provided here.
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Thread starter 2 weeks ago
#3
Sorry, the question begins with: "The marks of candidates in an examination are modelled by the continuous random variable X with p.d.f."
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Thread starter 2 weeks ago
#4
(Original post by RDKGames)
As far (iii) is concerned, I think it's fairly obvious what type of equation that is. I mean... if you expand the whole thing, what's the highest power you're going to have? Will it have a positive or negative coefficient? So how would it look like?? You can snipe out the roots of it too since it's in a factorised form already, so you should have a good idea of how it looks like.

For its second part of (iii), well clearly it is referring to some context which you haven't provided here.
What would i do for IV? And is a sufficient answer for the first part of III: The graph is a negative quartic with intercepts at (0,0), (50,0) and (100,0)?
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2 weeks ago
#5
(Original post by TAEuler)
What would i do for IV? And is a sufficient answer for the first part of III: The graph is a negative quartic with intercepts at (0,0), (50,0) and (100,0)?
Well you answers half of (III) so try to answer the second half.
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Thread starter 2 weeks ago
#6
(Original post by RDKGames)
Well you answers half of (III) so try to answer the second half.
The distribution of this negative quartic is the sum of two smaller distributions, one for high-scoring candidates and the other for low-scoring candidates

Now what do I do for part iv)?
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1 week ago
#7
(iv) is quite almost a philosophical question. It's often encountered in finance where anything tradable is going to be denominated in discrete units but modelled using continuous processes. The rationale for doing so is that you get a huge body of tools and techniques "for free" once you transfer the problem into the continuous world while discrete things are clunky and hard to generalise.I'm not sure what official answer is expected here. Perhaps you could argue that the underlying reality you're trying to model is continuous and the marks are merely appoximations to this anyway.
Last edited by SerBronn; 1 week ago
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1 week ago
#8
(Original post by SerBronn)
(iv) is quite almost a philosophical question. It's often encountered in finance where anything trade-able is going to be denominated in discrete units but modelled using continuous processes. The rationale for doing so is that you get a huge body of tools and techniques "for free" once you transfer the problem into the continuous world while discrete things are clunky and hard to generalise.I'm not sure what official answer is expected here. Perhaps you could argue that the underlying reality you're trying to model is continuous and the marks are merely appoximations to this anyway.
Surely it's both a philosophical and practical question. In the extreme case of trading a single unique item (e.g. a painting), an analysis assuming you can either have 0.5 paintings, or for that matter, 2 paintings, is going to be rather horribly off the mark. More realisitically, I'd expect different behaviour when trading something where the smallest quanta is worth 0.01p v.s. Â£1000000. [However, I know little about the finance industry].

In this case, in practical terms, relative to the other errors a model this crude is going to create, I'd argue continuity corrections are the least of your worries.

I would like to say that an examiner would be happy with either "no, not permissible, particularly because with only 100 discrete values the continuity corrections are too big to ignore" or "yes, it is permissible, because with 100 discrete values the continuity corrections will be negligable" (i.e. the examiner not caring particularly about the actual answer as long as there's some indication the candidate knows what the issue is and has thought about it), but sadly in these days of rigid markschemes I fear that wouldn't actually be the case.
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Thread starter 1 week ago
#9
(Original post by DFranklin)
Surely it's both a philosophical and practical question. In the extreme case of trading a single unique item (e.g. a painting), an analysis assuming you can either have 0.5 paintings, or for that matter, 2 paintings, is going to be rather horribly off the mark. More realisitically, I'd expect different behaviour when trading something where the smallest quanta is worth 0.01p v.s. Â£1000000. [However, I know little about the finance industry].

In this case, in practical terms, relative to the other errors a model this crude is going to create, I'd argue continuity corrections are the least of your worries.

I would like to say that an examiner would be happy with either "no, not permissible, particularly because with only 100 discrete values the continuity corrections are too big to ignore" or "yes, it is permissible, because with 100 discrete values the continuity corrections will be negligable" (i.e. the examiner not caring particularly about the actual answer as long as there's some indication the candidate knows what the issue is and has thought about it), but sadly in these days of rigid markschemes I fear that wouldn't actually be the case.
Unfortunately it specifies the correct answer is "yes, if the step size is small compared with the standard deviation." Could you explain to me the answer and why this works and how the standard deviation matters?
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1 week ago
#10
(Original post by TAEuler)
Unfortunately it specifies the correct answer is "yes, if the step size is small compared with the standard deviation." Could you explain to me the answer and why this works and how the standard deviation matters?

This is a diagram showing how the normal distribution approximates a (p=0.5) binomial. You can see that the more "steps" you have, the closer the step curve is to the continuous curve.

The standard deviation really just gives you a measure of the "scale" that you're working with. Suppose I divide all the scores by 10 so the range is now 0 to 10. If we still had a step size of 1 (e.g. we round the scores to the nearest integer), each step is going to "effectively" be 10 times bigger than our range was 0 to 100. So the step size that is "small enough" would actually go down by a factor of 10. Which is the same as the reduction in the standard deviation caused by that scaling.
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Thread starter 1 week ago
#11
(Original post by DFranklin)

This is a diagram showing how the normal distribution approximates a (p=0.5) binomial. You can see that the more "steps" you have, the closer the step curve is to the continuous curve.

The standard deviation really just gives you a measure of the "scale" that you're working with. Suppose I divide all the scores by 10 so the range is now 0 to 10. If we still had a step size of 1 (e.g. we round the scores to the nearest integer), each step is going to "effectively" be 10 times bigger than our range was 0 to 100. So the step size that is "small enough" would actually go down by a factor of 10. Which is the same as the reduction in the standard deviation caused by that scaling.
Okay, so what you're saying is:
The ability to reexpress continuous random variables is dependent on the step size of the discrete random variable. In my example, the step size is 1. This is a very small step size which means that the discrete curve becomes very close to the continuous curve?

But i'm not really sure how a small standard deviation relates to that? And what is classed as a small one?
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1 week ago
#12
(Original post by TAEuler)
Okay, so what you're saying is:
The ability to reexpress continuous random variables is dependent on the step size of the discrete random variable. In my example, the step size is 1. This is a very small step size which means that the discrete curve becomes very close to the continuous curve?

But i'm not really sure how a small standard deviation relates to that? And what is classed as a small one?
The point is that 1 is only small relative to the spread of the distribution (we have values going from 0 - 100). If the spread was only (0-1), obviously a step size of 1 would be pretty rubbish.

It's easier (but technically "wrong") to look at things the other way around - we've been talking about approximating a discrete distribution with a continuous one, but instead imagine approximating a continuous one with a discrete one.

To talk about something slightly more concrete; imagine approximating a semicircle with the squares on graph paper. If your semicircle is large relative to the squares on the graph paper, it's a good approximation - if the diameter is only a few times greater than the width of a square, the approximation is going to be poor.

The size of the squares corresponds to the 'step size'; the diameter corresponds to the standard deviation (recall the standard deviation is a measure of the "spread" of the distribution).

Note that "small relative to the standard deviation" is a bit of a judgement call. 20% of the standard deviation is obviously "not small", and 0.01% almost certainly does count as "small". But exactly what is "small enough" is (IMHO) a bit of a judgement call.

It's also worth noting that "relative to the standard deviation" is more a rule-of-thumb than a completely reliable measure. If you imagine a game where I toss a coin, and give you Â£90 if it's heads, Â£10 if it's tails, and then throw a dice 5 times and give you the total thrown in pennies, then the distribution will have 2 very sharp peaks 25 pence wide centered at Â£90.15 and Â£10.15. The standard deviation is going to be Â£20, but a step size of 20p isn't going to give you any kind of accurate feel for the distribution.
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Thread starter 1 week ago
#13
(Original post by DFranklin)
The point is that 1 is only small relative to the spread of the distribution (we have values going from 0 - 100). If the spread was only (0-1), obviously a step size of 1 would be pretty rubbish.

It's easier (but technically "wrong") to look at things the other way around - we've been talking about approximating a discrete distribution with a continuous one, but instead imagine approximating a continuous one with a discrete one.

To talk about something slightly more concrete; imagine approximating a semicircle with the squares on graph paper. If your semicircle is large relative to the squares on the graph paper, it's a good approximation - if the diameter is only a few times greater than the width of a square, the approximation is going to be poor.

The size of the squares corresponds to the 'step size'; the diameter corresponds to the standard deviation (recall the standard deviation is a measure of the "spread" of the distribution).

Note that "small relative to the standard deviation" is a bit of a judgement call. 20% of the standard deviation is obviously "not small", and 0.01% almost certainly does count as "small". But exactly what is "small enough" is (IMHO) a bit of a judgement call.

It's also worth noting that "relative to the standard deviation" is more a rule-of-thumb than a completely reliable measure. If you imagine a game where I toss a coin, and give you Â£90 if it's heads, Â£10 if it's tails, and then throw a dice 5 times and give you the total thrown in pennies, then the distribution will have 2 very sharp peaks 25 pence wide centered at Â£90.15 and Â£10.15. The standard deviation is going to be Â£20, but a step size of 20p isn't going to give you any kind of accurate feel for the distribution.
Okay, so: The step size of 1 is small relative to the spread of the distribution and the standard deviation, this means it is permissible to model the mark with a continuous random variable like X, as defined in the question. This is due to the fact that when the step size is small relative to the spread of the distribution and the standard deviation, the graphs of the discrete variable and continuous variable become very close

Is that a sufficient explaination??

Also, i'm not sure if it is in fact small compared to the standard deviation..although I don't need this for the answer itself, as you seem it is small compared to the standard deviation. But I seemed to have worked out the standard deviation, and it is not large compared to the step size of 1.

Workings:
E(X) = 50
E(X^2) = 25000/7
Var (X) = 7500/7
standard deviation = (50root21)/7 = 32.73268354...
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