for 2 indépendant uniform random variables X , Y in [0,3] find conditional pdf f(XY | X = 1/3)

To find the conditional probability density function (pdf) of XY given X = 1/3, we need to use the conditional probability formula and the properties of independent variables.

Given that X and Y are independent uniform random variables in the range [0,3], we can start by writing down their individual probability density functions.

First, let's find the marginal pdf of X:
fX(x) = 1/(b-a) = 1/(3-0) = 1/3, for x in [0,3]
where a = 0 and b = 3.

Since X and Y are independent, the joint pdf of X and Y is the product of their marginal pdfs:
fXY(x, y) = fX(x) * fY(y) = (1/3) * (1/3) = 1/9, for (x, y) in [0,3] x [0,3]

Now, let's find the conditional pdf of XY given X = 1/3, denoted as f(XY | X = 1/3):

fXY(x, y) 1/9, for (x, y) in [0,3] x [0,3]
fY|X(y | x = 1/3) = ---------------- = ------------------------------
fX(1/3) 1/3, for x = 1/3

To find the region over which we need to evaluate this conditional pdf, we need to consider the condition X = 1/3. Since X is fixed to the value 1/3, we can ignore the x in the joint pdf and only evaluate the y values within the range [0,3].

Hence, the conditional pdf f(XY | X = 1/3) becomes:
f(XY | X = 1/3) = fY|X(y | x = 1/3) = (1/9) / (1/3) = 1/3, for y in [0,3]

Therefore, the conditional pdf of XY given X = 1/3 is a constant distribution with the value 1/3 over the range [0,3].