WitCH 85: The Continuation of MAV’s Trials

With Methods exams next week, this one’s kinda important.

We try to avoid critiquing, or even being in the same room as, third party VCE practice exams. They are invariably clunky and weird, with plenty to criticise, but they matter infinitely less than the yearly screw-ups of the official exams.

Even MAV trial exams we do our best to ignore. Yes, the MAV is (too) closely aligned with the VCAA (with a number of people in conflicted, dual roles), and so MAV has a significantly greater professional and moral obligation to maintain high standards. But still, third party is third party, and we try our best to just ignore MAV’s nonsense. On occasion, however, MAV’s nonsense matters sufficiently, or is simply sufficiently annoying, to warrant a whack.

Last week, we were shown of a copy of MAV’s 2022 Trial Exam 2 for Mathematical Methods. The exam is predictably bad, simultaneously reflective of a thoroughly perverted subject and eccentrically third party. The exam is CAS-obsessed, both in question structure and in suggested solutions. To indicate finding the extrema of the function \boldsymbol{f(x) = \sin^2 x \cos x} with a couple CAS screenshots, and without a single word noting that the critical points can also easily be found by hand, is so dismissive of the notion of teaching that we cannot conceive of what purpose the authors imagine there is in any of this. The exam is also replete with absurd modelling and Magrittisms. But none of that is why we are here.

One particular question on the MAV trial exam was pointed out to us. Here it is, together with MAV’s solution:

Get to work: students have exams next week.

UPDATE (30/10/22)

Thank you to everyone for the comments. It wasn’t hard, and was really more a PoSWW than a WitCH. But there seems to be some common misunderstanding of these issues, and we wanted to encourage the discussion. Here is a summary, and a couple more comments.

The first and least important thing to note is that the question is screwed, even on its own terms. The authors’ intention was to make the pdf s(x) continuous by doing the standard endpoint matching stuff, but it is impossible to do so. That is, the parameters a and b cannot be chosen to make s(x) have integral 1 and be continuous at both x = 14 and x = 16. (Two parameters and three equations, so unless you’re really lucky …)

The second and most important thing to note is that the authors’ assumption that the pdf s(x) should be continuous is completely and utterly false. Continuity is simply not required for a function to be a probability density function. Ever. It is not required even if one chants the magic words “continuous random variable”. The use of “continuous” in CRV refers to the continuity of the cumulative distribution function, \boldsymbol{S(x) =\int\limits^x_{-\infty} s(t)\,{\rm d} t}, which will be true as long as the integral of s(x) makes sense in some standard manner.*

The third and unimportant thing to note is that one might argue on the basis of a given model that the CDF S(x) should be differentiable. If s(x) is continuous then that will be true, with S'(x) = s(x) continuous.** But one must argue for this, the trial exam authors don’t argue for this, and their “model” of cats sleeping is so absurd it would be impossible to argue for this.

*) In some more advanced courses the CDF of a CRV need not be continuous. See the brief discussion of MCQ14 here (in which VCAA stuffed up and then attempted to hide their stuff up). See also VCAA’s definition of “continuous variable” (Word, idiots) and ACARA’s definition of “continuous random variable” and ACARA’s other, contradictory definition of “continuous random variable”, and ACARA’s v9 definition of “continuous numerical variable” (Word, idiots).

**) Don’t say S(x) is “smooth”. The word doesn’t mean what VCAA thinks it means. 

UPDATE (30/10/22)

John Friend and Sai have noted in comments that the question can be reasonably fixed by asking for the set of all a-b combos that make s(x) into a pdf.

52 Replies to “WitCH 85: The Continuation of MAV’s Trials”

  1. I wonder who writes these exams, and who proofreads them… this is such a simple issue that it could be solved by just adding the word “continuous” before probability density function. Or maybe MAV thinks that all probability density functions must be continuous. Or that it’s somehow “implied” by the context.

    That’s not to mention that the problem is completely meaningless and trivialised anyways by the use of the CAS. And the completely meaningless context. Though I suppose it’s good preparation for the VCAA exams, which are also 90% CAS and 10% parsing inane and stupid contexts.

    1. The probability density function would not be continuous using the intended a and b anyway, between the middle piece and the bottom piece around x=16.

      1. Good point. I guess this problem is completely screwed then. Either way this problem should have been sorted out by proofreading.

        1. Thanks for this very important post, Marty.

          The relationship between the MAV and the VCAA has been too cosy (as detailed by Marty) for too long a time. And with the MAV providing presentations by VCAA staff (behind a paywall) on the new Mathematics Study Design, MAV’s influence on teachers is particularly powerful at the moment. So not only is the problem “completely screwed”, but the error has the potential to ‘teach’ something that is very wrong (to both students and teachers). It is what I call an influential and dangerous error.

          It is interesting to note that the MAV website asserts that
          “MAV commissions an experienced author team to prepare the papers and has instituted a three-stage review process to ensure that each paper covers the course appropriately and is free of errors.”
          It’s reasonable to “wonder who writes these exams, and who proofreads them”. It’s also interesting to speculate on why the “three-stage review process” failed to detect this influential and dangerous error. I have some theories but am interested in what others think (*). And I wonder if this error is being taught or implicitly abetted in some classrooms?

          From a pragmatic point of view, I also wonder if the pdf is used in subsequent parts of the question. If so, I can see unwanted difficulties in figuring out consequential marks (particularly for students who are smart enough to say that the question is defective and that it’s impossible to find a and b!)

          On the subject of ‘fixing’ this question, the question simply needs to ask “Find all possible values of a and b”.
          By solving

          \displaystyle \int_{14}^{16} a(x-16)^2 + b \, \, dx = \frac{1}{2}

          under the restrictions \displaystyle a \geq 0 and \displaystyle b \geq 0 (applying the often overlooked criterion that a pdf cannot be negative, easy to do with a CAS) you get

          \displaystyle 0 \leq a \leq \frac{3}{16} and \displaystyle b = \frac{3 - 16a}{12}.

          Any subsequent part using the pdf can then give a particular valid value for a and b in its preamble (“For Burmese cats it is found that a = … and b = … Find …) and there is no issue with consequential marks.

          For the students who read this post, I would like to make it crystal clear that there is no necessity for the pdf of a continuous random variable to be continuous. The only criteria are:

          \displaystyle \int_{-\infty}^{\infty} f(x) dx = 1, and

          \displaystyle f(x) \geq 0.

          NEVER assume continuity of the pdf unless it is specifically stated in the question. As Tungsten correctly points out, although the MAV solution incorrectly assumes continuity of the pdf, it’s not continuous “using the intended a and b anyway, between the middle piece and the bottom piece around x=16”. Hoist with its own petard.

          * Could this strengthen the argument for including more mathematical statistics such as the cumulative distributive function (cdf) and deleting the inferential statistics crap?

          PS – Don’t get me started on the gratuitous use of brackets between the integral and the dx in the first two lines of the MAV ‘solution’. This is always guaranteed to get my idiot detector tingling.

  2. They skipped thinking about equating b=0 if they were really “trying” to make the function continuous. I wonder if the person who wrote this forgot that the continuous uniform distribution existed… The “solution” accounts for one case but as long as you pick a and b such that the PDF is non-negative through \frac{8 a}{3}+2 b+\frac{1}{2}=1 then that’s also equally as valid. I think the actual solution should be b \in [0,\frac{3}{8}] and a =\frac{1}{16}(3-12b). Seems like a really fine line between being stupidly hard and being illogical for no good reason.

  3. As a matter of interest – what are students in Mathematical Methods expected to know about continuous functions?

      1. Thanks Marty. I did not find any mention of continuous functions in the study design. Does the text book deal with continuous functions?

        1. The standard SD expression is “informal consideration of continuity”. The topic appears in the current Cambridge 34 text in Chapter 9 (after differentiation …).

            1. Golly gosh, TM. You should know that “Informal consideration” is VCAA code for “We don’t want to do this rigorously because it might scare teachers and students”.

              Mathematical rigour is anathema to VCAA (and many teachers too, these days).

    1. I know some teachers who do teach \displaystyle \lim_{x \rightarrow a} f(x) = f(a) but not too many. This is certainly not in the Study Design (but it is in Cambridge) and I don’t know why. I think the most that many students get is a “You can trace your pen along the graph without lifting it off” style treatment.

      1. But then Terry has a point. Does “both sides” stuff ever come up in VCAA exams? If not, then why is it an MAV trial exam?

  4. Suppose that the upcoming exam has a function defined and asks if this is a valid probability density function. And suppose a student who has trained on this trial exam decides “no” because the function is not continuous. Will they be able to sue the MAV?

      1. Marty, you’ve accidently quoted the VCAA disclaimer.

        Anyway, it would certainly be an interesting test case and potentially re-shape the landscape. I’ve never seen a disclaimer on \displaystyle any trial exam, let alone a MAV exam. But I do know that writers and vettors sign a contract and I would imagine that somewhere in that contract there was a clause that said that vettors must ensure that the trial exams are free of mathematical errors … Maybe there’s a disclaimer somewhere in the purchase agreement …?

        The VCAA Examination Report very wisely always states the following:
        “Note: This report provides sample answers, or an indication of what answers may have included. Unless otherwise stated, these are not intended to be exemplary or complete responses.”

  5. Thanks for the update, Marty. I really think this is one of your more important posts and will be of great value to students (and teachers). I will be giving every Maths Methods student I teach a link to this post. The MAV has actually done the teaching community a great (albeit unintentional) service with this question by uncovering what is possibly a very common misunderstanding. (*)

    Re: Update point 1. Even if the writers do want to argue that the continuity at x = 14 is reasonable:
    a) The argument has to be made (and, as you note, it was not), and
    b) Additionally, the argument also has to *ahem* argue that continuity at x = 14 is reasonable but continuity at x = 16 is NOT. I’d love to see \displaystyle that part of the argument.

    At the risk of sounding immodest (a risk I take daily), a way of fixing the question has been proposed, at no charge. Not only does it fix the error, it improves the question in terms of potential consequential marks for later parts. A two-pronged fix. Should the MAV read your post.

    * And as Barnum once said, there’s no such thing as bad publicity!

    1. Thanks, JF. Yes I should have included your (and Sai’s) suggestion on fixing the problem. I’ll update the update when I’m back at the computer.

  6. The cdf of a discrete random variable can (obviously) be discontinuous. And the cdf of a random variable that is a mixture of discrete and continuous can be discontinuous. However, I don’t believe that the cdf of a purely continuous random variable can be discontinuous

    I thought that a real valued random variable has a pdf if and only if its cdf is absolutely continuous (a stronger condition than continuity). So …

    Re: Update and important point 2 * and the random variable defined here: https://www.qedcat.com/methods2014.html

    “However, a continuous random variable so defined may still take on particular values with strictly positive probability. (Consider for example tossing a coin, setting X = 0 if the toss is heads, and otherwise sampling X randomly from a normal distribution.)”

    I’m not sure that this random variable is purely continuous (or are we entering into the abstract realm of measure theory?)

    1. In fact, I’m sure it would be considered neither discrete nor continuous. Anyway, it’s only a small thing for my own curiosity (some Summer time reading).

    2. Well you added the word “purely”. We’re just nitpicking definitions. But note the definitions of ACARA and VCAA.

      1. Yes, I read their definitions.

        I’ve actually found it tricky to define a continuous random variable for Maths Methods students. One possibility is the following:

        X is a continuous random variable if there is a nonnegative function f(x), called the probability density function of X (or pdf), such that

        \displaystyle \Pr(X \in B) = \int_{B} f(x) \, dx

        for every subset \displaystyle B of the real line and \displaystyle  \int_{-\infty}^{\infty} f(x) \, dx = 1.

        In particular, the probability that X falls within an interval is \displaystyle \Pr(a < X < b) = \int_{a}^{b} f(x) \, dx.

        And then give examples. Or the other way around … But this seems like a big overkill …? I 'confess' to using the following:

        Continuous random variable: A random variable that can take on any value in an interval or a union of disjoint intervals of the real number line.

        To others (including Marty) – What definition would/do you give? I'm happy to be criticised and led to a better definition (keeping in mind that the definition needs to be at the Maths Methods level and (mostly *) correct).

        * Some level of compromise might be required …?

        1. Yes, it is tricky. But the first thing is to understand it’s tricky, that the intuitive definition of “can take all values” doesn’t mean what people think it means. It doesn’t imply the existence of a pdf. In VCE I’d be happy with the intuitive definition standardly given and then note the *assumption* that we’ll always have a pdf.

          1. With the risk of sounding contrarian, I don’t think that the assumption that there’s always a PDF is good enough.

            You give the example of “consider for example tossing a coin, setting X = 0 if the toss is heads, and otherwise sampling X randomly from a normal distribution”, and while that application may be contrived, it’s pretty easy to come up with a similar example that is somewhat realistic, like the battery life of a battery when fully charged: there’s an above zero that the battery life would be 0, due to the battery being dead, so there’s no PDF for that function. Or another would be the price of an option at its expiry date: there’s an above zero chance that the option would be worth $0.

            1. Thanks, Anonymous. Nice example.

              I didn’t mean to imply there really is always a pdf. Just that if that is the context VCE wishes to restrict to, it needs to state that in some explicit manner, and not implicitly assume it follows from “takes all values”.

            2. Hi Anon. (It’s hard to keep a track of you all!).

              I think if we’re dealing with a random variable that is ‘purely’ continuous, the assumption of it having a pdf is good enough. Certainly within the scope of Maths Methods, where the random variables are either discrete or continuous but not a mixture of both.

              Yes, there are ‘mixed random variables’ met in higher level tertiary mathematical probability subjects and actuarial studies. Indeed, in such cases, there is no pdf. Instead, you have a pdf and a pmf over different regions. But they \displaystyle still have a cdf (*). Such is the case with your more realistic example, I think.

              * A major deficiency of the Mathematics Study Design, in my opinion, is the absence of cdf’s and their use. A bit too mathematical for VCAA, which prefers to stick with its inferential crap.

        2. JF,

          Trouble is the statistics in Methods is somewhat neutered…

          For those interested in statistics I find it helpful to emphasize that a pdf is a “probability density function” whereas a CDF is a “Cumulative Distribution Function” and distinguish between discrete and continuous distributions. So the probabilities will always lie between zero and one and the CDF will sum or integrate to 1 over the entire range.

          Reading any introductory statistics book such as Kreyszig (1970) would also be helpful
          (but maybe not a day before the exam)

          Stack exchange had a go explaining the relationship in simple terms


          Steve R

          1. Showing your age with that reference, Steve! Kreyszig (Advanced Engineering Mathematics) is a great textbook. I have a slightly later edition (which nevertheless still shows \displaystyle my age!)

            A continuous random variable (and its pdf, when it exists) is defined from the cdf. Things are much ‘simpler’ and clearer. Unfortunately the cdf is not part of the Mathematics Study Design – too mathematical for VCAA, I suppose. It was briefly on the Yr 12 curriculum in the late 80’s (as an option for Mathematics B – 12 Topic 3: Continuous probability distributions) under VCAB (1986 – 91), but obviously got the old heave after the take-over by VBOS, BOS and then VCAA diluted things.

            1. Correction/clarification: But a continuous cdf does not always define a pdf (and hence a continuous random variable). The Cantor distribution (see earlier post) is a counter-example.

              1. Thanks, Terry. The main point with VCE is for the Bosses to note that their intuitive definition of continuous RV (“takes on all values”) does not imply what they really want (probabilities are given by integrating).

                It’s not that hard. Which means it’s too hard for ACARA and VCAA.

            2. John,

              Your correct about age . Kreyszig’s engineering text is a classic and still used at Monash today

              I also like his introduction to statistical methods which has a number of practical examples if you can
              find a reseller

              Steve R

  7. One easy fix for the question is to simply present it along with a graph of the intended s(x), showing that it is indeed intended to be continuous at x = 14. Perhaps that’s how the question was originally written, but then some … at MAV decided that it was “too easy” with the graph presented and removed it, not realising that it broke the question.

    The context of sleeping cats is still absolutely ridiculous though.

    1. Language … (I edited your comment slightly).

      I’m not thrilled with “copy the picture” questions. In this case, I think it would make it worse.

      1. That reminds me of an another question I saw on a practice exam by a much smaller company than MAV. The question asked something along the lines of “give all stationary points” of a function f within a certain domain. The solution listed a certain amount of stationary points, which they got by using the local maxima/minima feature on the graph application on the CAS. It turned out that there was one extra turning point, that was too small and too close to the endpoints to be seen directly from the graph, so they didn’t notice it.

    2. Hi edderiofer.

      I think you give the MAV too much credit. Or perhaps too little … Because a graph that appears to show continuity at x = 14 does not mean that there is continuity at x = 14 (https://mathematicalcrap.com/2017/10/25/the-treachery-of-images/)
      and should not be included in the question.

      And even if it was and we suspend our belief, how is no continuity at x = 16 justified?

      No, I don’t think a graph was ever included. But I can imagine a discussion where the error was raised but then people got gas lighted into believing that continuity was reasonable, never realising that there cannot be continuity everywhere.

      1. Or maybe they thought that continuity inside the support of the pdf was more reasonable/important (or perhaps even required) than outside … (yep, grasping straws here).

        Then again … I haven’t seen too many textbook questions where there’s a discontinuity inside the support. So maybe some people have only ever seen white sheep and soon develop the (mistaken) belief that these are the only sheep that exist …? (This could explain why some people might think continuity at x = 14 is required but not at x = 16).

        1. Related to this, I have not seen any textbook questions of the type
          “Find the value of k so that f(x) defines a probability density function”
          where there are several values of k arising from solving \int_{-\infty}^{\infty} f(x) \, dx = 1 but some must be rejected because of f(x) \geq 0 (the often overlooked, unsexy second criterion).

          I’m never surprised when I give my class such a question. See attached for an example.

  8. Nice question. The 0<x may trick some students as well… (in exam stress, it is possible that k=0 may be disregarded and k=1 is given as the only solution (when, as you say, this should be rejected)).

    99% (or more) of these “find k” questions seem to have k as only a scale factor, so the “above the axis” criterion is never really a consideration. Art imitating life? Textbooks trying to predict VCAA’s thinking?

    1. A good textbook won’t aid and abet ‘all sheep are white’ thinking.

      Same thing happens with horizontal asymptotes in Maths Methods – students never see an example where the graph crosses a horizontal asymptote and so construct the wrong idea. Often assisted by teachers who glibly and incorrectly state that a graph can never intersect an asymptote.

      Ditto “Find all solutions to \displaystyle f(x) = f^{-1}(x)” questions when \displaystyle f(x) = x doesn’t work. Plenty of trial exams (AND VCAA) have had an opportunity to provide a more careful, thoughtful solution (and perhaps even include a brief discussion) but have not (in some cases, refused point blank upon request – they know who they are).

      Ditto differentiability at a point when the derivative is not continuous at the point.

      The list goes on and on and on like a Status Quo song (but not nearly as enjoyable). And we wonder why get stuck with such a crud curriculum.

      We have an exam error list (which is a huge timesaver for addressing many student questions at this time of year. Thankyou, Marty! I encourage all teachers to give the links to their students). Maybe we need a list of all the misconceptions that VCAA and its flunkies encourage among students and teachers.

  9. The distribution function of a rv X is defined by F_{X}(t) := P(X \leq t) for t \in R.

    We say that X is a continuous rv if its distribution function is continuous.

    See Shiryaev, A.N. (1989). Probability. (Second ed.) Springer, (p. 171)

    1. Hi Terry. The word ‘continuous’ in the context of a function is the trouble. As we know, there is continuity (the sort of continuity dealt with in Maths Methods), uniform continuity and absolute continuity (which is stronger than continuity and uniform continuity).

      So technically ‘we’ must say that X is a continuous random variable if its cdf is \displaystyle absolutely continuous.

      For example, the Cantor distribution (https://en.wikipedia.org/wiki/Cantor_distribution) is continuous but does not define a continuous random variable. It defines a \displaystyle singular random variable (neither discrete or continuous or a mixture).

      Obviously ‘we’ can’t say all of this in Maths Methods, but I think it can be hinted at enough to say that the given definition of a continuous random variable in Maths Methods comes with certain assumptions.

      I think many Maths Methods students would be fascinated to know that there are other types of random variables apart from discrete and continuous. I think they would be even more fascinated to know that a pdf is not required to calculate expected value and variance – these calculations can be done using only the cdf. It’s a genuine shame the the cdf is not in the Study Design (at least for Specialist Maths if not Maths Methods).

      The cdf of a random variable would be a terrific theme for a Specialist Maths SAC 3 (but unfortunately this idea is trumped by the fact that it is far more pragmatic for SAC 3 to focus on exam style probability and statistics questions).

      1. I should point out that a proof of the formulae for finding the expected value and variance from the cdf makes (cunning) use of integration by parts (on the new Study Design) but involves improper integrals (not on the new Study Design).

      2. Yes, the different uses of the word “continuity” can be an issue.

        According to Shiryaev, there is a difference between a continuous random variable and an absolutely continuous random variable. He defines a continuous random variable as above.

        Uniform continuity applies to functions generally rather than random variables.

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