The Student Room Group

STEP Maths I, II, III 1989 solutions

Scroll to see replies

Reply 20
STEP II, Q5

i) x=rcosθx = r\cos\theta
y=rsinθy = r\sin\theta

dydx=dydθ/dxdθ\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}\theta} / \frac{\mathrm{d}x}{\mathrm{d}\theta}

Alternatively (refer to the first attachment):

tanϕ=1rdrdθ\displaystyle \tan \phi = \frac{1}{r} \frac{\mathrm{d}r}{\mathrm{d}\theta}

dydx=tanx=tan(90o(ϕθ))=cot(ϕθ)=1+tanθtanϕtanϕtanθ\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \tan x = \tan (90^{o} - (\phi - \theta)) = \cot(\phi - \theta) = \frac{1 + \tan\theta \tan\phi}{\tan\phi - \tan \theta}

=1+1rdrdθtanθ1rdrdθtanθ=drdθtanθ+rdrdθrtanθ\displaystyle = \frac{1 + \frac{1}{r} \frac{\mathrm{d}r}{\mathrm{d}\theta} \tan\theta }{\frac{1}{r}\frac{\mathrm{d}r}{\mathrm{d}\theta} - \tan\theta} = \frac{\frac{\mathrm{d}r}{\mathrm{d}\theta} \tan \theta + r}{\frac{\mathrm{d}r}{\mathrm{d}\theta} - r\tan\theta}

ii) Refer to the second attachment.

dydx=tan(90oθ2)\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = - \tan(90^{o}-\frac{\theta}{2})

Equating this equal to the above expression gives

Unparseable latex formula:

\displaystyle \frac{\mathrm{d}r}{\mathrm{d}\theta} = - r \frac{1 - \tan\theta \tan(90^{o}-\frac{\theta}{2})}{\tan\theta + \tan(90^{o}-\frac{\theta}{2})} = - r\cot(90^{o} + \frac{\theta}{2}}) = r\tan\frac{\theta}{2}



Solving the differential equation by separation of variables gives

r=Asec2θ2=Acos2θ2=2A1+cosθ=a1+cosθ\displaystyle r = A \sec^{2}\frac{\theta}{2} = \frac{A}{\cos^{2}\frac{\theta}{2}} = \frac{2A}{1 + \cos \theta} = \frac{a}{1 + \cos\theta}

This is the polar form of a parabola.
Reply 21
*Bump* :biggrin:
Reply 22
I'd quite like help with STEP I question 16.

For part 2, I get E(J) = b_1 + 2b_2 + 3b_3 and we're trying to find b_1, b_2 and b_3 to maximise E(J) under the condition that b_1 + b_2 + b_3 = 1. I know I'm meant to somehow make use of the fact that player A chooses his numbers with probability 1/3. And using our answer from part 1, we know what B should choose given that A has chosen a certain number.

I realise I've just rewritten the question pretty much but yeah... a push in the right direction would be nice.

On a side note, it's very depressing to spend like an hour doing a question (STEP II question 7) and not getting it.
Your expression for E(J) seems way off. How did you get it?

But suppose it was right. Are you saying you don't know how to maximize b1+2b2+3b3b_1 + 2b_2 + 3b_3 given b1+b2+b3=1b_1+b_2+b_3 = 1, or that you don't know how to justify it? (I don't think any justification is needed).
Reply 24
Well, unless I'm misunderstanding, its PDF is given by

j123P(J=j)b1b2b3\begin{array}{r|lcc} j & 1 & 2 & 3 \\ \hline P(J=j) & b_1 & b_2 & b_3 \end{array}

And as E(X)=xP(X=x)E(X) = \sum xP(X=x)

You get E(J) = b_1 + 2b_2 + 3b_3

And I'm saying I don't know how to maximise that. All I can think of is

E(J) = 1 + b_2 + 2b_3

But then I'm stuck. I guess that b_2 = 0 and b_3 = 1 would maximise it though, but it's only a guess.
Hmmm - I thought you'd just chosen poor notation.

You shouldn't be looking at E(J) as far as I can see. You're supposed to be finding the expected winnings.

As far as the maximization goes: you're close, but you've forgotten about b_1.
Reply 26
Oh yes, I see your point. Here's another attempt then

Suppose A picks 1;

If B picks 1, he makes 0. Associated probability = b1 / 3
If B picks 2, he makes -1. Associated probability = b2 / 3
If B picks 3, he makes 2. Associated probability = b3 / 3

Expected winnings when A picks 1 = 2b3/3b2/32b_3/3 - b_2/3

Suppose A picks 2;

If B picks 1, he makes 1. Probability = b1 / 3
If B picks 2, he makes 0. Probability = b2 / 3
If B picks 3, he makes -1. Probability = b3 / 3

Expected winnings when A picks 2 = b1/3b3/3b_1/3 - b_3/3

Suppose A picks 3;

If B picks 1, he makes -2. Probability = b1 / 3
If B picks 2, he makes 1. Probability = b2 / 3
If B picks 3, he makes 0. Probability = b3 / 3

Expected winnings when A picks 3 = b2/32b1/3b_2/3 - 2b_1/3

So expected winnings = E(A picks 1) + E(A picks 2) + E(A picks 3)

= 2b3/3b2/3+b1/3b3/3+b2/32b1/32b_3/3 - b_2/3 + b_1/3 - b_3/3 + b_2/3 - 2b_1/3

= b3/3b1/3b_3/3 - b_1/3

Is that right? And surely this is maximised when b_3 = 1, b_2 = 0 and b_1 = 0?
To me, the more "natural" way of doing this would be the other way around. Calc the expected winnings when B plays 1, then the expected winnings when B plays 2, etc. But it doesn't really matter, and your way is probably going to work better for (iii).

In essence what you've done is right, but you're not doing the conditional expectations properly. All your "expected winnings when A picks 1 (or 2, or 3)" calculations are too small by a factor of 3. But your final answer ends up being right, because

E(X)=E(XY)P(Y)E(X) = \sum E(X | Y)P(Y), which means the 1/3 factor appears here.

[What you've done isn't exactly a serious error - it's more you've got the notation wrong than the actual mathematics. But it is wrong].

I think you've done all you need to for the maximization though.
Reply 28
DFranklin

In essence what you've done is right, but you're not doing the conditional expectations properly. All your "expected winnings when A picks 1 (or 2, or 3)" calculations are too small by a factor of 3.


Just for clarification, is this the proper way to do it then?

P(B picks 1|A picks 1) = P(B and A pick 1)/P(A picks 1) = 3P(B and A pick 1) = 3 * P(B picks 1) * 1/3 = b_1.

But then where does the 1/3 factor appear? I see your formula but I don't understand it - why do we multiply by P(Y)?

On a side note, it's quite an interesting result. Might just program something to test it out over 1,000,000 iterations or something.
Swayum
Just for clarification, is this the proper way to do it then?

P(B picks 1|A picks 1) = P(B and A pick 1)/P(A picks 1) = 3P(B and A pick 1) = 3 * P(B picks 1) * 1/3 = b_1.Why are you working this out in the first place? (What you've done is almost certainly "correct but tautological". You are assuming A and B are independent in your calculation, in which case you know P(B | A) = P(B) already).
Reply 30
Oh yeah, that's true. I'm still confused about where the E(X) = sigma(E(X|Y)P(Y)) formula comes from though.
The basic definition of expectation is E(X)=xiP(X=xi)E(X)=\sum x_i P(X=x_i).

But p(X=xi)=jP(X=xiY=yj)P(Y=yj)p(X=x_i) = \sum_j P(X=x_i | Y=y_j) P(Y=y_j) (where y_j goes over all possible values of Y).

So E(X)=i,jxiP(X=XiY=yj)P(Y=yj)E(X) = \sum_{i,j} x_i P(X=X_i | Y=y_j) P(Y=y_j)

=j[ixiP(X=XiY=yj)]P(Y=yj)=\sum_j \left[\sum_i x_i P(X=X_i | Y=y_j)\right] P(Y=y_j)

=jE(XY=yj)P(Y=yj)=\sum_j E(X|Y = y_j) P(Y=y_j)

Does your textbook actually cover conditional expectation? Hopefully there's a longer (better) explanation there.
Reply 32
Nope, I don't think it does. S1 just defines it as E(X) = sigma(xP(X=x)) and mentions stuff about E(aX + b). S2 makes a shift from discreet random variables to continuous random variables, which essentially just replaces the sigma with an integral sign. Haven't done S3 onwards. The syllabus was probably just different in 89. I'll try and make sense of the above tomorrow when I'm fresh :p:. Thanks a lot for your help.
Well, you don't have to do the problem using conditional expectation. But that's what it looks like you're doing when you write "Expected winnings when A picks 1 = 2b_3/3 - b_2/3". (If I rewrite as "Expected winnings given A picks 1 = 2b_3/3 - b_2/3" it should be more obvious it's basically a conditional calculation).
I wouldn't sweat it too much - you're basically messing up the notation a bit; my trying to clarify is clearly not helping, so just don't worry about it. It won't cost you more than a mark.

The neatest way (and I mean literally, in the sense of the clearest way of showing what you're doing) of approaching this is probably to write two matrices:

a payoff matrix showing who wins what for each of the 9 possible pairs (a,b),

and a probability matrix showing the probability of each of the 9 possible pairs (a,b).

Then you can just write down the expectation.
for the older STEP papers, could put the links up on the wiki: http://www.thestudentroom.co.uk/wiki/Mathematics_Past_Papers
the links on the Big Fat Step Megathread dont work.
Thanks
Reply 36
the_mathematical_fish
for the older STEP papers, could put the links up on the wiki: http://www.thestudentroom.co.uk/wiki/Mathematics_Past_Papers
the links on the Big Fat Step Megathread dont work.
Thanks


This link (which is in the megathread) has every paper ever published; http://www.byond.com/members/DeathAwaitsU/files/STEP.zip .
Reply 37
DFranklin
...


Right so for the 3rd part, I get

Expected winnings = (2b3b2)a1+(b1b3)a2+(b22b1)a3(2b_3 - b_2)a_1 + (b_1 - b_3)a_2 + (b_2 - 2b_1)a_3

(So X = 2b_3 - b_2, Y = b_1 - b_3, Z = b_2 - 2b_1)

And for the next part, is it enough if I just state

b_1 = 0.25
b_2 = 0.5
b_3 = 0.25?

With these numbers, X, Y and Z all turn out 0 which is non-negative. And so the expectation is 0 as well, regardless of a_1, a_2 and a_3, which means that he won't LOSE any money (in the "long run").
Swayum
This link (which is in the megathread) has every paper ever published; http://www.byond.com/members/DeathAwaitsU/files/STEP.zip .


Wow, thanks Swayum!
Hokay heres my attempt at STEP II/9 can get everything out part from the last bit which is annoying because i feel like i've covered something like this or theres something I should know but its lost in the ether.

STEP II Question 9 Partial

A2=(3183)(3183)=(1001) A^2 = \begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}\begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

F=I+n=11n!tnAn F = I + \displaystyle \sum_{n=1}^{\infty} \frac{1}{n!} t^n A^n

F=I+tA+12t2A2+16t3A3+14!t4A4....... F = I + tA + \frac{1}{2}t^2 A^2 + \frac{1}{6}t^3 A^3 + \frac{1}{4!}t^4 A^4.......

Now first thing considering the matrices A^3,A^4 and etc..

A = A^3 = A^5 = etc....

A^2 = A^4 = A^6 etc... = I

so this

F=I+tA+12t2A2+16t3A3+14!t4A4 F = I + tA + \frac{1}{2}t^2 A^2 + \frac{1}{6}t^3 A^3 + \frac{1}{4!}t^4 A^4

becomes

F=I+tA+12t2I+16t3A+14!t4I.... F = I + tA + \frac{1}{2}t^2 I + \frac{1}{6}t^3 A + \frac{1}{4!}t^4 I....

F=I(1+12t2+14!t4....)+A(t+16t3....) F = I(1 + \frac{1}{2}t^2 + \frac{1}{4!}t^4....) + A(t + \frac{1}{6}t^3....)

Now looking at the taylor series for sinht and cosht (which I'll exclude from this). Its obvious cosht=1+12t2+14!t4... cosht = 1 + \frac{1}{2}t^2 + \frac{1}{4!}t^4... and sinht=t+16t3.... sinht = t + \frac{1}{6}t^3.... .

So

F=Icosht+Asinht F = Icosht + Asinht

Now to find F1 F^{-1} we note that

FF1=I=A2 FF^{-1} = I = A^2

F1=A2Icosht+Asinht F^{-1} = \frac{A^2}{Icosht + Asinht}

=A2(IcoshtAsinht)(Icosht+Asinht)(IcoshtAsinht) = \frac{A^2(Icosht - Asinht)}{(Icosht +Asinht)(Icosht - Asinht)}

=A2(IcoshtAsinht)I2cosh2tA2sinht = \frac{A^2(Icosht - Asinht)}{I^2cosh^2t - A^2sinht}

=I(IcoshtAsinht)I=IcoshtAsinht = \frac{I(Icosht - Asinht)}{I} = Icosht - Asinht

F1=IcoshtAsinht F^{-1} = Icosht - Asinht

Finally

dFdt=Isinht+Acosht \frac{dF}{dt} = Isinht + Acosht

FA=AIcosht+A2sinht=Isinht+Acosht FA = AI cosht + A^2 sinht = Isinht + Acosht

so

dFdt=FA \frac{dF}{dt} = FA

Last part im stuck on

drdt+Ar=0 \frac{dr}{dt} + Ar = 0

Integrating factor matrix = eAdt=eAt e^{\int A dt} = e^{At}

eAtdrdt+AeAtr=0 e^{At}\frac{dr}{dt} + Ae^{At}r = 0

ddt(reAt)=0 \frac{d}{dt}(re^{At}) = 0

reAt=C re^{At} = C

r=CeAt r = Ce^{-At}

Now I've read something that would help here but i can't put my finger on it.

How do i go from this to finding x(t) and y(t) ?

Quick Reply

Latest