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?
No one has answered this question yet.
It is a simple problem, but if you are not at ease with conditional density functions one soon gets confused.
The first thing you need to do is to introduce a second (discrete) random variable, Y, that represents the choice of the machine. So Y can take on values 1 and 2 and only these values.
Possible density functions involved are f(x,y), f1(x), f2(y), g1(x|y) and g2(y|x).
We know g1(x|y) and f2(y):
f2(y) = 0.5 for y=1,2 (machines are chosen with equal probability).
g1(x|y) = 2x if y=1, and 3x^2 if y=2 (condition on machine chosen).
Now, part (a) asks f1(x) [function associated with probability that X=x] which we can compute with g1 and f2; part (b) asks g2(y|x) [function associated with probability that Y=y, given that X=x] which we can compute with g1, f2, and the in part (a) found f1.
a) f1(x) = sum over all values of y expression g1(x|y)f2(y) = 2x*0.5 + 3x^2*0.5.
b) g2(y|x) = with Bayes' law for p.d.f.'s g1(x|y)f2(y)/f1(x). We know all three functions, so we are near an answer.
g2(y|x) = g1(x|1)f2(1)/f1(x) if y=1, and g1(x|2)f2(2)/f1(x) if y=2 = 2/(2+3x) if y=1, and 3x/(2+3x) if y=2.
So P(Y=1|X=0.25) = 2/(2+3x) = 2/(2+3*0.25) = 8/11.
To determine the marginal probability density function (p.d.f.) of X, we need to calculate the probability density of X for each instrument and then combine them.
(a) Marginal p.d.f. of X:
Let's denote the random variable representing the choice of instrument as B. Since either instrument 1 or instrument 2 is chosen at random, we have P(B = 1) = P(B = 2) = 1/2.
To find the marginal p.d.f. of X, we need to consider the joint probability density function (p.d.f.) of B and X. We can express this as:
f(x) = P(X = x) = P(X = x, B = 1) + P(X = x, B = 2)
Since B can take two values (1 or 2), we can rewrite the above equation as:
f(x) = P(X = x | B = 1) * P(B = 1) + P(X = x | B = 2) * P(B = 2)
Using the given p.d.f.s:
f(x) = (2x * 1/2) + (3x^2 * 1/2)
= x + 3/2x^2, for 0 < x < 1
Therefore, the marginal p.d.f. of X is f(x) = x + 3/2x^2, for 0 < x < 1.
(b) Probability that instrument 1 was used given X = 1/4:
To find the probability that instrument 1 was used given X = 1/4, we need to calculate P(B = 1 | X = 1/4).
Using Bayes' theorem:
P(B = 1 | X = 1/4) = (P(X = 1/4 | B = 1) * P(B = 1)) / P(X = 1/4)
We can calculate the individual probabilities:
P(X = 1/4 | B = 1) = f1(1/4) = 2(1/4) = 1/2
P(B = 1) = 1/2
P(X = 1/4) = ∫[0,1] (x + 3/2x^2) dx
Evaluating the integral:
P(X = 1/4) = ∫[0,1] (x + 3/2x^2) dx
= [(1/2)x^2 + (1/2)(3/2)x^3] from 0 to 1
= (1/2)(1^2 + (3/2)(1^3)) - (1/2)(0^2 + (3/2)(0^3))
= 1/2 + (3/4)
= 5/4
Substituting the values:
P(B = 1 | X = 1/4) = (1/2 * 1/2) / (5/4)
= 1/4 / (5/4)
= 1/5
Therefore, the probability that instrument 1 was used given X = 1/4 is 1/5.