Transformation of Random Variables

Suppose X is a random variable with PDF fX and CDF FX. Let Y = r(X) be a function of X, then we call Y = r(X) a transformation of X.

In discrete case, the mass function of Y is given by
fY(y) = P(Y = y) = P(r(X) = y) = P({x; r(x) = y}) = P(X ∈ r-1(y))

Example: Let us take the following mass function –

xfX(x)
-11/4
01/2
11/4

Let Y = X2. Then, P(Y = 0) = P(X = 0) = 1/2 and P(Y = 1) = P(X = 1) + P(X = -1) = 1/2. Summarizing:

yfY(y)
01/2
11/2

For continuous case, there are three steps for transformation
1. For each For each y, find the set Ay = {x : r(x) ≤ y}.
2. Find the CDF
                     Fy(y) = P(Y ≤ y) = P(r(X) ≤ y)
                               = P({x; r(x) ≤ y}) = ∫Ay fx(x)dx
3. The PDF is fY(y) = F’Y(y).

Example: Let fX(x) = e−x for x > 0.
Hence, FX(x) = ∫0x fX(s)ds = 1 − e−x …(i)
Let Y = r(X) = logX …(ii)
Then, Ay = {x : x ≤ ey} and
            FY(y) = P(Y ≤ y) = P(logX ≤ y) …using eqn(ii)
            = P(X ≤ ey) = FX(ey) = 1 − e−ey … using eqn(i)
Therefore, fY(y) = eye−ey for y ∈ R.

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