Final answer:
The PDF of X, a continuous random variable uniformly drawn given Y=y, is determined by integrating the conditional density over Y's range. It results in fX(x) = ln(4) - ln(max(3, x)), for 0 ≤ x ≤ 3, showing the density function varies based on Y's interval [3, 4].
Step-by-step explanation:
When considering a uniform distribution, the probability density function (PDF) for a continuous random variable X, given that X is conditioned on another uniform random variable Y, can be found by understanding the definition of a uniform distribution. Since Y ~ Uniform([3, 4]) and given Y = y, X is drawn uniformly from [0, y], the PDF of X is 1/y for x between 0 and y. This is because the area under the PDF must equal 1, and for a uniform distribution, the density function is constant across the interval.Furthermore, because Y is uniformly distributed between 3 and 4, the marginal PDF for X across the combined intervals determined by Y is the average of the PDFs conditioned on Y. Thus, to obtain fX(x), we need to integrate the conditional PDF of X from 3 to 4 with respect to Y. The PDF of X therefore equals:
fX(x) = ∫34 (1/y) dy, for x in [0, 3],
which simplifies to:
fX(x) = ∫max(3,x)4 (1/y) dy,
this is because when x > 3, X cannot have a value greater than its upper limit, which is Y. Carrying out the integration, we get:
fX(x) = ln(4) - ln(max(3, x)), for 0 ≤ x ≤ 3.
This resulting function shows that X has a uniform distribution with a density function that changes depending on the value of x relative to Y's range [3, 4].