## Now you need a maths degree to decide if you get PIP

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- Category: Latest news
- Created: Thursday, 07 January 2016 12:54

A recent personal independence payment (PIP) decision by an upper tribunal says that probability theory must be used to calculate eligibility for claimants who have more than one health condition. The decision even goes into great lengths about how to mathematically calculate the probability of two independent conditions occurring for more than 50% of the time.

**Bad days**

The bizarre decision relates to a claimant who had both chronic obstructive pulmonary disease (“COPD”) and rheumatoid arthritis.

The first tier tribunal found that the claimant had 3 bad days a week for COPD, and 2 to 3 bad days a week, depending on the weather, for rheumatoid arthritis. Each condition affected the claimant’s ability to carry out some of the activities in the PIP test, such as preparing food or dressing.

The tribunal decided that the claimant’s conditions did not cause him to satisfy any point scoring descriptors for over 50% of the time, as required by regulation 7, so he was not eligible for PIP.

**Probability theory**

The upper tribunal judge, however, held that the tribunal had failed to find out if the effect of one condition was enough to cause problems or whether there was only an issue when both conditions applied at the same time.

If the effects of either condition on its own was enough to allow points to be scored, it would still be necessary to work out how many days a week on average, the claimant was affected by one or both conditions. If 50% or less then no award of PIP could be made.

Relying on a submission by the DWP, the tribunal judge went on to set out in great detail how to calculate the probability of the two conditions occurring for more than 50% of the time.

We have reproduced, some, though by no means all, of the mathematical evidence below.

*If we say that the rheumatoid arthritis occurs 3 times a week then the probability that regulation 7 is satisfied is worked out as follows: *

*Probability(A or B) = Probability(A) + Probability(B) – Probability(A and B)*

*The probability of event A (arthritis) occurring, 3/7, is added to the probability of event B (COPD) occurring. 3/7 + 3/7 =6/7. *

*However, this probability (6/7) also includes the times on which the COPD and arthritis both occur. This must be subtracted. It is double counting. *

*The probability of the COPD and arthritis both occurring is 3/7 x 3/7 = 9/49.*

*We need to subtract 9/49 from 6/7. To do this we convert 6/7 into 42/49 (by multiplying by 7). We can then do the simple sum 42/49 - 9/49 = 33/49. This is the probability of an operative condition being present on a given day.*

*33/49 can otherwise be understood as 4.71/7, if you were looking at this on a weekly basis (as we have been here). *

*Therefore the descriptor is satisfied for more than 50% of the days.*

**Help!**

If there are any maths bods out there who can draw us up a table covering other possible combinations where no condition causes 4 or more bad days a week – what about when a claimant has three conditions, for example – we’ll be happy to publish it.

You can read the full decision here: [UK/972/2015]

## Comments

If a, b, c, and d, represents the number of GOOD days a week for each condition (treat c and d as 7 if only 2 conditions are present, then the probable bad days a week should be: -

(1 – ((a/7) * (b/7) * (c/7) * (d/7))) * 7

In the example given in UK/972/2015, the two conditions each occurred 3 days a week, so each condition has 4 good days a week.

(1 – ((4/7) * (4/7))) * 7 = 4.71 (approx)

I think the minimum requirements are (assuming the benchmark is more than 3.5 days a week), one of the following: -

A condition with 4 or more bad days a week (obviously)

A condition with 3, and a second condition with 1 bad day a week.

(1 – ((4/7) * (6/7))) * 7 = 3.57 (approx)

Three conditions (2 2 1)

(1 – ((5/7) * (5/7) * (6/7))) * 7 = 3.94 (approx)

Four conditions (2 1 1 1)

(1 – ((5/7) * (6/7) * (6/7) * (6/7))) * 7 = 3.85 (approx)

Or 5 conditions, each with one bad day a week.

(1 – ((6/7) ^ 5)) * 7 = 3.76 (approx)

I have to say that this is one of the most ridiculous and ill thought out UT decisions I have seen. It raises the question, what if a claimant has four conditions. This totals 16 different combinations of conditions per day. Does a tribunal have to make a decision on whether or not each individual combination affects Regulation 7, and then do the mind-boggling calculation of probabilities?

That said, if a claimant can keep a diary of how his or her conditions affect day-to-day activity, this should provide far stronger factual evidence, and reduce the need for probability theory.

I learned an important lesson that day. Statistics do not go by "common-sense" rules and those who don't know stats should stay clear of making pronouncements about the maths involved.

By making this decision and at the level where it becomes a formal precedent, the UT judge has left it open for any claimant who can find a willing mathematician to argue whether the DM, any LT tribunal and any previous UT have got their stats wrong.

I would add one other thing which I learned about stats, which is that complicated as it is when you have 2 independent factors to calculate, it's worse still if you have to work out a probability that the two factors are NOT independent. In this case if the COPD is made worse when the RA flares up - or vice versa. Or not.

However, the DWP are not going to love this judge either. All of a sudden any DM's judgement can be questioned if s/he doesn't have a post-graduate diploma in stats.

So I'm guessing that the DWP will shortly issue a

fatwalaying out the rules for multiple conditions - they might even make an opportunity out of a crisis and declare that the 50% can only be calculated on one condition. Then get that through the Commons by threatening to make the MPs argue the stats and hope there aren't too many retired statisticians in the Lords.Lets take an urn containing the ashes of claimant killed off by IDS, and put in seven balls in 3 red and 4 white, take a second urn containing the ashes of a victim of a miscarriage of justice who can't prove his innocence and put in 4 blue balls and three white balls. The first decision is whether this approximates to the real life situation we are analysing when we draw one ball from each urn; the second decision is whether we replace the balls after each draw.Don't get the ashes mixed up, the relatives won't like it. The correct method and model to use is not in the A level syllabus: the Bernoulli,binom ial,normal,expo nential, Chi-square distributions are there but not the relevent multinomial and hypergeometric distributions presumably these will be met in the first year of a numerate degree. The explanation for the method requires several large gulps of post A level pure mathematics; don't swallow them all at once it will really hurt. The judge did not just get it plain wrong he demonstrated that he was ignorant of the theory of probability and could never have understood why he got wrong because he lacked the necessary mathematical skills, in his overconfidence he decided to become his own expert and inflicted damage on the claimant. I have no space to provide a calculation but might try again. The reason I like statistics are the wonderful square roots, the gorgeous factorials and Iove listening to the horrid screams.

I don't quite get the same answer as Gordon for three conditions. If one applies two days, the others one day, so 2+1+1 in his notation then the probabilities of good days are, respectively, 5/7, 6/7, 6/7. Multiplying these gives you 180/343 which is over a half. If two of the conditions apply on two days (2+2+1), though, you do qualify.

It does all seem a bit artificial doesn't it? Things get even worse if the conditions are not independant as you'd need to somehow assign various conditional probabilities (how?!) and then use Bayes' Theorem.

For three conditions the only set of values that is not over 50% is 1+1+1.

It does all seem a bit absurd: reminds me of those wretched boxes on the old DLA forms where I had to say exactly how many minutes I needed help with something, even if I wasn't getting it! Still the tribunal ruling is case law now so this sort of thing could come up in the future, so I thought I'd try to clarify the maths, for those who felt up to getting to grips with it.

Imagine you are the claimant, you have these two unpleasant conditions and when you wake in the morning you wonder if you are going to have a good day. For this to happen you have to not be significantly affected either by your COPD or your arthritis. Since your COPD affects you three days a week you have a four out of seven chance of not being affected. Ditto for the arthritis.

So the probability of a good day is the chance of not having bad COPD (4/7) times the chance of not having bad arthritis (also 4/7). This gives 16/49 which is just under a third. So only about one in three days are good days and so about two out of three are bad, clearly qualifying you for PIP under the 'descriptor applies 50% of the time' rule.

According to my calculations the more general situation for such a pair of conditions is this: if one applies at least three days a week then, even if the other only affects you one day a week you will still have bad days more than 50% of the time. If they both affect you only two days a week then, unfortunately it works out to just a whisker below the 50% threshold.

Hope this helps ... perhaps someone could check I've done my maths right and comment if I'm mistaken.