Let the random variables and have joint pdf
Find the conditional pdf .

Question

Let the random variables  and  have joint pdf  
Find the conditional pdf .

BlackTom AI · Accepted Answer

We start by identifying the conditional pdf f_{Y|X}(y|x) = f(x,y) / f_X(x) for the given joint pdf f(x,y) = 6y with the support 0 < y < x < 1.
First, compute the marginal f_X(x) by integrating f(x,y) over y from 0 to x: f_X(x) = ∫_{0}^{x} 6y dy = 3x^2, valid for 0 < x < 1.
Next, form the conditional: f_{Y|X}(y|x) = f(x,y) / f_X(x) = (6y) / (3x^2) = 2y / x^2.
This expression is valid for the same support as f(x,y): 0 < y < x < 1. Outside this region, the conditional pdf is 0 since the joint pdf is 0 there.
Now evaluate the given options:
- Option 1: 2y/x^2, 0 < y < x < 1. This matches the derived form exactly and respects the support. It appears correct.
- Option 2: -1/(1-y), 0 < y < x < 1. This is dimensionally inconsistent with the derived f_{Y|X}(y|x) and does not come from the integral or ratio, so it is incorrect.
- Option 3: 2y, 0 < y < x < 1. This lacks the necessary division by x^2; it does not incorporate the x-dependence of the conditional density, so it is incorrect.
- Option 4: (2y)/(x^2), 0 < y < x < 1. This is essentially the same as Option 1 but written with the fraction in a slightly different notation; since the content is identical, it represents a correct form as well. If the option set treats it as distinct, it would still be equivalent to the correct expression.
In summary, the correct conditional pdf is the one that equals 2y/x^2 over the region 0 < y < x < 1, and zero elsewhere. The other options either omit the necessary x-dependence, alter the functional form, or present an expression not consistent with the joint-to-marginal ratio.

AMS 310 Homework 7

Let the random variables and have joint pdf Find the conditional pdf .

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