Suppose that either of two instruments might be used for making a certain measurement. Instrument 1 yields a measurement whose p.d.f. is

f1(x)=2x, 0 <x<1

Instrument 2 yields a measurement whose p.d.f. is
f2(x)=3x^2, 0 <x<1

Suppose that one of the two instruments is chosen at random and a measurement X is made with it.
(a)
Determine the marginal p.d.f. of X.
(b)
If X = 1/4 what is the probability that instrument 1 was used?

To determine the marginal probability density function (PDF) of X, we need to consider the probabilities of selecting each instrument and the corresponding PDFs.

(a) Determining the marginal PDF of X:
The marginal PDF of X, denoted as fX(x), is the overall PDF of X regardless of which instrument was used.

We need to consider the probabilities of selecting each instrument (in this case, both equally likely) and the corresponding PDFs:

Let P(I1) be the probability of selecting instrument 1, and P(I2) be the probability of selecting instrument 2. Since they are chosen at random, P(I1) = P(I2) = 1/2.

Now, let's calculate the marginal PDF of X. We can do this by taking the weighted sum of the PDFs of X for each instrument:

fX(x) = P(I1) * f1(x) + P(I2) * f2(x)

Substituting the values, we have:

fX(x) = (1/2) * (2x) + (1/2) * (3x^2)
= x + (3/2) * x^2

Therefore, the marginal PDF of X is given by:

fX(x) = x + (3/2) * x^2, 0 < x < 1

(b) To find the probability that instrument 1 was used when X = 1/4, we need to use Bayes' theorem.

Let P(I1|X = 1/4) be the probability that instrument 1 was used given X = 1/4. Using Bayes' theorem, we have:

P(I1|X = 1/4) = (P(X = 1/4|I1) * P(I1)) / P(X = 1/4)

First, let's calculate P(X = 1/4|I1), the probability that X = 1/4 given instrument 1 was used. We can find this by substituting X = 1/4 in f1(x):

P(X = 1/4|I1) = f1(1/4)
= 2 * (1/4)
= 1/2

Next, we need to calculate P(X = 1/4), the probability that X equals 1/4. We can do this by integrating the marginal PDF of X over the interval [1/4, 1/4]:

P(X = 1/4) = ∫[1/4,1/4] (x + (3/2) * x^2) dx
= ∫[1/4,1/4] (x) dx + ∫[1/4,1/4] ((3/2) * x^2) dx
= (1/4) * (1/4) + (3/2) * (1/4)^3
= 1/16 + 3/32
= 5/32

Now, substituting the known values back into Bayes' theorem:

P(I1|X = 1/4) = (P(X = 1/4|I1) * P(I1)) / P(X = 1/4)
= (1/2 * 1/2) / (5/32)
= 16/20
= 4/5

Therefore, the probability that instrument 1 was used when X = 1/4 is 4/5 or 0.8.