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Assume we have a Gamma Random Variable $X$ with the following pdf

$$ \frac{m^mx^{m-1}}{\Gamma(m)}\text{exp}(-mx)$$

If I am asked to find the distribution of the following $$Y= aX$$ where a is non-negative constant, how will the distribution change?

Thanks

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1 Answer 1

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If $Y=aX$ then it follows that $$\begin{align} f_Y(y) & = f_{aX}(y) \\[1ex] & = \frac{f_X(y/a) }{ a } \end{align}$$


Note: This is analogous to $\mathsf P(Y=y) = \mathsf P(aX=y) = \mathsf P(X=y/a)$ for a discrete RV; though scaling a probability density distribution is slightly different from scaling a probability mass distribution.


A gamma distribution has two parameters, shape $\alpha$ and rate $\beta$.

$$Z\sim \operatorname{Gamma}(\alpha, \beta) \iff f_Z(z) = \dfrac{\beta^\alpha z^{\alpha - 1} e^{-\beta z}}{\Gamma (\alpha)}$$

So...

$$\begin{align} f_X(x) & = \dfrac{m^m x^{m - 1} e^{-m x}}{\Gamma (m)} \\\ \therefore X & \sim \operatorname{Gamma}(m, m) \\[3ex] f_Y(y) & = \frac{f_X(y/a)}{a} \\[1ex] & = \dfrac{1}{a} \dfrac{m^m a^{1-m} y^{m - 1} e^{-m y /a}}{\Gamma (m)} \\[1ex] & = \dfrac{(m/a)^m y^{m - 1} e^{-(m/a) y}}{\Gamma (m)} \\[2ex] \therefore Y &\sim \operatorname{Gamma}(m, m/a) \end{align}$$

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