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?
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To determine the marginal probability density function (pdf) of X, we need to consider the probability of each instrument being chosen and the corresponding pdf of each instrument.
(a) To find the marginal pdf of X, we first need to calculate the probability of each instrument being chosen. Since there are only two instruments and they are chosen at random, each instrument has an equal probability of 1/2.
For Instrument 1, the pdf is given as f1(x) = 2x, where 0 < x < 1.
For Instrument 2, the pdf is given as f2(x) = 3x^2, where 0 < x < 1.
To find the marginal pdf of X, we need to calculate the weighted sum of the individual pdfs of each instrument, considering their probabilities. Since both instruments have the same probability (1/2), the marginal pdf of X, let's call it fX(x), is given by:
fX(x) = (1/2) * f1(x) + (1/2) * f2(x)
Substituting the given pdfs of instruments 1 and 2, we have:
fX(x) = (1/2) * 2x + (1/2) * 3x^2
Simplifying this equation, we obtain:
fX(x) = x + (3/2)x^2, where 0 < x < 1.
Therefore, the marginal pdf of X is fX(x) = x + (3/2)x^2, where 0 < x < 1.
(b) To find the probability that instrument 1 was used given X = 1/4, we need to use Bayes' theorem.
Let A be the event that instrument 1 was used, and B be the event that X = 1/4.
We need to find P(A|B), the probability of instrument 1 being used given X = 1/4. Bayes' theorem states:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability of observing X = 1/4 given that instrument 1 was used. In this case, it is simply f1(1/4) = 2 * (1/4) = 1/2.
P(A) is the probability of instrument 1 being chosen, which is 1/2.
P(B) is the probability of observing X = 1/4, which can be calculated by integrating the marginal pdf of X from 1/4 to 1:
P(B) = ∫[1/4 to 1] fX(x) dx
Substituting the equation for fX(x) we obtained earlier and integrating, we have:
P(B) = ∫[1/4 to 1] (x + (3/2)x^2) dx
Integrating the above expression, we find:
P(B) = [1/2 x^2 + (1/2)x^3] evaluated from 1/4 to 1
P(B) = [(1/2 * 1^2 + (1/2) * 1^3) - (1/2 * (1/4)^2 + (1/2) * (1/4)^3)]
P(B) = [1/2 + 1/2 - 1/32 - 1/128]
P(B) = 31/32
Now we can calculate P(A|B) using Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (1/2 * 1/2) / (31/32)
Simplifying, we get:
P(A|B) = 16/31
Therefore, the probability that instrument 1 was used given X = 1/4 is 16/31.