Consider a random sample $(X_1, \ldots, X_n)$ where all $X_i$ are iid r.v. with the following density function: $$f(x, \theta) = Cx^{-(p+1)} e^{-\theta/x } 1_{[0,+\infty)}(x)$$ Find the value of $C$ and prove that $U_i = \frac{1}{X_i}$ has the gamma distribution, determine its shape, mean and variance.

I have already found $C$ integrating pdf (seems like $C = \frac{1}{\Gamma(p)}\theta^p$)

My question: How distribution of a rv $X$ is related to $\frac{1}{X}$ and what theory is behind it ? I will appreciate any kind of hints and paper/books recommendations.

  • 1
    $\begingroup$ Your normalization constant should be positive. Otherwise this integral is easily reduced to Gamma integral by changing variable to $y=1/x$. Or you could simply look it up in Gradshtein&Ryzhik. $\endgroup$
    – Roger V.
    Commented Dec 18, 2019 at 12:48

2 Answers 2


First of all, $C$ cannot be negative, because for non-negative $x$ the $x^{-(p+1)} $ and $e^{-\dfrac{\theta}{x} }$ are non-negative, which means that integral also should be non-negative. You should get $C = \dfrac{1}{\Gamma(p)}\theta^p$ (probably you forgot to change integration bounds after transforming $dx$ to $d(\frac{1}{x})$)

Second, in order to get the prove, you need to apply formula that defines relations between two distributions when the $f$ function is being used as a transformation that converts one random variable to the other (based on these notes: https://www.math.arizona.edu/~jwatkins/f-transform.pdf).

$$ f_Y(y) = f_X(g^{-1}(y))\left|\frac{d g^{-1}(y)}{dy}\right| = f_X\left(\frac{1}{y}\right) \, \frac{1}{y^2} $$

where $y = g(x) = \frac{1}{x}$ and $x = g^{-1}(y) = \frac{1}{y}$

And now you just need to expand previous statement, so that you get

$$ f_Y(y) = f_X\left(\frac{1}{y}\right) \, \frac{1}{y^2} = y^{p-1} \, e^{-\theta y} $$

It's a density function of the unnormalised gamma distribution (see https://en.wikipedia.org/wiki/Gamma_distribution). It's easy to find the right coefficient in order to covert it to the proper PDF (you've done it already).

And the last part related to the mean and the variance. It's easy to calculate these values. Mean could be solved by solving the following integral (in the same way you did it for the $C$ coefficient)

$$ E[y] = \int_0^{\infty} y \, p_Y(y) dy $$

and similar for variance

$$ Var(Y) = E[y^2] - E[y]^2 $$

Where you need to find

$$ E[y^2] = \int_0^{\infty} y^2 \, p_Y(y) dy $$

While solving these integrals you should take advantage of the following identity

$$ \Gamma(z) = \frac {\Gamma(z + 1)} {z} $$


For a one-dimensional distribution: if you have $y=g(x)$, where $g(x)$ is a monotonic function with inverse $x = g^{-1}(y)$, and $x$ is distributed with the probability density $f_x(x)$, then the probability density of $y$ is \begin{equation} f_y(y) = f_x(g^{-1}(y))\left|\frac{d g^{-1}(y)}{dy}\right| = f_x(g^{-1}(y))/\left|\frac{d g(x)}{dx}\right| \end{equation}

The proof is easy, if you consider the normalization of the probability: \begin{equation} 1 = \int_0^{+\infty}dx f_x(x) = \int_0^{+\infty} dy\frac{dx}{dy} f_x(x(y)) = \int_0^{+\infty} dy f_x(g^{-1}(y))\left|\frac{d g^{-1}(y)}{dy}\right| = \int_0^{+\infty}dy f_y(y) \end{equation}

  • $\begingroup$ There has to be some problem with the proof. In the beginning, you've mentioned that $g(x)$ has to be monotonic, but proof is being applied to any $g$ even if there is no one to one mapping. For example, let's say $g(0) = 1$ and $g(2) = 1$ then there is no way of reconstructing x from the y value: $g^{-1}(1)=0$ or $g^{-1}(1)=2$ $\endgroup$
    – itdxer
    Commented Dec 18, 2019 at 16:29
  • $\begingroup$ You are right, there have to be one-to-one mapping, it has to be differentiable, etc. So just monotonicity is not enough. $\endgroup$
    – Roger V.
    Commented Dec 18, 2019 at 16:43
  • $\begingroup$ Also, I have another question, how did you get from the step 3 to the step 4 (place where absolute value appeared)? $\endgroup$
    – itdxer
    Commented Dec 18, 2019 at 16:53
  • $\begingroup$ Your function $g(x)$ is either monotonously increasing or monotonously decreasing. If it is increasing than its derivative is positive and equals to its absolute value. If it is decreasing then its derivative is negative, but you also need to swap the limits of integration. If you go yourself through the derivation for $y = g(x) = 1/x$ you will see how it works ;) $\endgroup$
    – Roger V.
    Commented Dec 18, 2019 at 17:03

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