What is the conditional probability density function 𝑓𝑌|𝑋(𝑦|𝑥) of 𝑌 given 𝑋=𝑥, if 𝑋 is uniformly distributed on the interval [𝑎,𝑏] and 𝑌 is uniformly distributed on the interval [𝑘,𝑧]?
To find the conditional probability density function fY|X(y|x) of Y given X=x, we need to use the concept of conditional probability. In this case, we have two continuous random variables, X and Y, where X is uniformly distributed on the interval [a, b] and Y is uniformly distributed on the interval [k, z].
The conditional probability density function fY|X(y|x) is defined as the probability density function of Y given that X takes the value x.
To derive the conditional probability density function, we can start by writing down the joint probability density function fXY(x, y) of X and Y. Since X and Y are independent variables, their joint probability density function is equal to the product of their individual probability density functions.
For X, which is uniformly distributed on the interval [a, b], the probability density function fX(x) is given by:
fX(x) = 1 / (b - a), for x ∈ [a, b],
= 0, otherwise.
Similarly, for Y, which is uniformly distributed on the interval [k, z], the probability density function fY(y) is given by:
fY(y) = 1 / (z - k), for y ∈ [k, z],
= 0, otherwise.
Since X and Y are independent, the joint probability density function fXY(x, y) is the product of their individual probability density functions:
fXY(x, y) = fX(x) * fY(y)
= 1 / (b - a) * 1 / (z - k), for x ∈ [a, b] and y ∈ [k, z],
= 0, otherwise.
To find the conditional probability density function fY|X(y|x), we need to normalize the joint probability density function by dividing it by the marginal probability density function of X, which in this case is fX(x). This normalization ensures that the conditional probability density function integrates to 1.
Thus, the conditional probability density function fY|X(y|x) is given by:
fY|X(y|x) = fXY(x, y) / fX(x)
= (1 / (b - a) * 1 / (z - k)) / (1 / (b - a)), for x ∈ [a, b] and y ∈ [k, z],
= 1 / (z - k), for y ∈ [k, z] and x ∈ [a, b],
= 0, otherwise.
So, the conditional probability density function fY|X(y|x) of Y given X=x is a constant 1 / (z - k) for y ∈ [k, z] and x ∈ [a, b], and it is 0 outside this region.
This means that given X=x, the distribution of Y is uniformly distributed on the interval [k, z].