• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

chi-square interpolation

K

kartik_newpro

Member
I have 2 queries regarding using interpolation to calculate chi-square probabilities. Assume chi-square = X for convenience

1. P(X4 > 1.064)

The chi-square percentage point tables directly shows this as 90%. But if I interpolate using the probability tables I get a different answer. Just tell me if interpolation is incorrect.

P(X < 1.0) = 0.0902 and P(X < 1.5) = 0.1734

0.0902 + (1.5 - 1.064)/(1.5 - 1.0) * (0.1734 - 0.0902) = 0.1628

P(X4 > 1.064) = 1 - 0.1628 = 0.8372 or 83.72%.

Where have I gone wrong?


2. P(X29 < 56.84)

From the tables, P(X > 52.34) = 0.05 and P(X > 58.30) = 0.01

Therefore,

0.01 + (58.30 - 56.84)/(58.30 - 52.34) * (0.05 - 0.01)

This doesnt give the right answer. What is the right way to interpolate?

Thanks
 
Just came across one more doubt and did not want to create a new thread for it.

Suppose X has a Poisson distribution and Y has a geometric distribution and X and Y are independent, then

Is P[X - Y] = P[X=x] - P[Y=y]?

Awaiting reply for both the doubts.

Hope I am not getting on the nerves of people here. I just ask too many doubts :p
 
Hi

You've made a mistake in your interpolation in your first query.

It should be:

0.0902 + (1.064 -1)/(1.5 - 1.0) * (0.1734 - 0.0902) = 0.1008

In your second query, your interpolation is fine, but you've used the wrong probability.

P(X29 > 52.34) = 0.005 and P(X29 > 58.30) = 0.001.

You might try drawing a line and labelling the end points with the two x values and the two probabilities. You can then put in the point you know, say the x point that lies somewhere between the two end points. You are trying to determine the unknown probability by splitting the difference in the probabilities in the same ratio as the x point splits the difference in the x values.
 
In response to your other query about P(X-Y) ...

No, you can't say P(X-Y) = P(X=x) - P(Y=y).

You will need to specify the exact values for X-Y and work out the ways this could happen. For example, if X = score on a die and Y is the number of heads when a coin is tossed twice, then X can take values 1,2,3,4,5,6 and Y can take values 0,1,2.

So X-Y could take on the values -1,0,1,2,3,4,5,6.

P(X-Y=-1) = P(X=1 and Y=2) = P(X=1) x P(Y=2) since they're independent, ie 1/6 x 1/4. The probability that X-Y takes on some values is more complicated because the value can happen in more than one way. For example, P(X-Y=0) = P(X=1 and Y=1) + P(X=2 and Y=2) etc.
 
Thanks a lot Margaret. That was super helpful.

About the second query - actually that was concerning a question in the May 2011 Exam. The question says -

X ~ Poi(2) and Y has geometric distribution with mean 2 where y = 0,1,2,..... Also X and Y are independent. Prove:

P[X=Y] = 1/3 * exp(-2/3)

I am not aware of any method other than finding P[X-Y=0]. But couldn't do it anyway :eek:

Can you help me on this?
 
In general

P(X=Y) = sum over all a>=0 { P(X=a | Y=a) * P (Y=a) } ...(1)

since X,Y independent

P(X=a|Y=a) = P(X=a)

Plug in the constants provided to get a sum over all a of
(e^-2 * 2^a /a!) * ((1/3)*(2/3)^a )

or

e^-2 (1/3) * sum over all a { (4/3)^a /a!} ....(2)

now e^x = sum over all a>=0 {x^a /a!} Taylor series

so the second bit of (2) is simply e^(4/3)

Thus (2) becomes
(1/3) e^-2 e^(4/3) = (1/3) e^(-2/3)


Alternatively (if you're not familiar with taylor series) you could note that each term of the second bit of 2 is e^(4/3) times the pdf of a Poisson dist with mu = 4/3; factor it out and sum the probabilities to get 1 and same results

the PDF of X-Y is a bit more complicated.
basically go down a similar route

P(X-Y=b) = sum over all a {P(X=a+b | Y=a) * P(Y=a)}
 
Thanks didster. I understood that well.

But E[Y] = 2. And the mean of geometric distribution is 1/p.

Then how come the PF is (1/3)*(2/3)^a ?

Shouldn't p be equal to 1/2 ?
 
Last edited by a moderator:
You could just use the stats function in your calculator. I find it a lot quicker... :cool:
 
Back
Top