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September 16, 2014

September 16, 2014

Posted by **Salman** on Thursday, December 2, 2010 at 1:17pm.

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.

- Stats and Data -
**Gerard**, Tuesday, March 15, 2011 at 6:27amIt 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.

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