Random Number Generation - What is the 'middle one of three'?

[Pyroadept]

Homework Statement

The logistic random variable X has CDF:

F(X) = exp(a + bx)/(a + exp(a + bx)), b>0

(i) Obtain a formula for random values of X in terms of R ~ U(0,1)

(ii) Using the cdf, suppose that a certain application requires you to generate 'the middle one of three' i.i.d. logistic - generate two such values.

Use the following sequence of random numbers. If you run out, start again from the beginning:

0.9892, 0.2842, 0.9349, 0.9029, 0.3494, 0.9891, 0.6072, 0.4893

Homework Equations

The Attempt at a Solution

Hi everyone,

Here's my solution so far:
--
(i)
Solve for X using R = F(X)

Multiplying across and factorising, we get:

exp(a+bx).(1 - R) = R

Now taking natural logs and regarranging, we get:

X = (1/b).[ln(R/(1-R)) - a]

... an equation for X in terms of R, so could call it X(R)

(ii)
I will generate random values of X using the following method:

Let random value of X = X(R_i),

where R_i is the ith random number given in the sequence above.

My question is - what do they mean by 'middle one of three'? Do I generate three random numbers and then pick the middle-valued one? It seems a bit straightforward, but I can't see what else it could be.

If that is the case, I get:

X_1 = (4.517-a)/b

X_2 = (-0.9237-a)/b

X_3 = (2.6645-a)/b


And, as b>0, the middle one is then X_2.

Is this correct?

Thanks for any help!

 

[mighty2000]

Hi there,
Yes, your solution for part (i) is correct.

For part (ii), you are on the right track. Yes, you generate three random numbers using the given sequence and then pick the middle one. So in this case, the middle one would be X_2. However, I think there is a mistake in your calculations for X_1 and X_2. It should be:

X_1 = (4.517 - a)/b
X_2 = (2.6645 - a)/b
X_3 = (-0.9237 - a)/b

Then, the middle one would be X_1.

I hope this helps!