If X is Pareto distributed and Y = ln(1 + X/theta), what distribution does Y have?

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Multiple Choice

If X is Pareto distributed and Y = ln(1 + X/theta), what distribution does Y have?

When you apply a monotone transformation to a random variable, you find the distribution of the new variable by transforming the CDF. If X is Pareto with scale x_m and shape alpha, its CDF is F_X(x) = 1 - (x_m/x)^alpha for x ≥ x_m. Define Y = ln(1 + X/θ). The mapping from Y to X is X = θ(e^y − 1), which is strictly increasing, so the support of Y starts at the point where X hits x_m: y_min = ln(1 + x_m/θ). The CDF of Y is F_Y(y) = P(Y ≤ y) = P(X ≤ θ(e^y − 1)) = F_X(θ(e^y − 1)) for y ≥ y_min, and F_Y(y) = 0 for y < y_min. Substituting the Pareto CDF gives F_Y(y) = 1 − [x_m / (θ(e^y − 1))]^α for y ≥ y_min. Differentiating gives the PDF f_Y(y) = α x_m^α / θ^α · e^y / (e^y − 1)^{α+1}, for y ≥ y_min. This distribution is not a standard Pareto, Normal, Exponential, or Gamma in general. Its tail behaves like an exponential in y (roughly proportional to e^{−αy} for large y), but the exact form involves (e^y − 1) in the denominator, so it does not match a pure Pareto or the other named families.

When you apply a monotone transformation to a random variable, you find the distribution of the new variable by transforming the CDF. If X is Pareto with scale x_m and shape alpha, its CDF is F_X(x) = 1 - (x_m/x)^alpha for x ≥ x_m. Define Y = ln(1 + X/θ). The mapping from Y to X is X = θ(e^y − 1), which is strictly increasing, so the support of Y starts at the point where X hits x_m: y_min = ln(1 + x_m/θ).

The CDF of Y is

F_Y(y) = P(Y ≤ y) = P(X ≤ θ(e^y − 1)) = F_X(θ(e^y − 1)) for y ≥ y_min,

and F_Y(y) = 0 for y < y_min. Substituting the Pareto CDF gives

F_Y(y) = 1 − [x_m / (θ(e^y − 1))]^α for y ≥ y_min.

Differentiating gives the PDF

f_Y(y) = α x_m^α / θ^α · e^y / (e^y − 1)^{α+1}, for y ≥ y_min.

This distribution is not a standard Pareto, Normal, Exponential, or Gamma in general. Its tail behaves like an exponential in y (roughly proportional to e^{−αy} for large y), but the exact form involves (e^y − 1) in the denominator, so it does not match a pure Pareto or the other named families.

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