1
$\begingroup$

Let $f_X (x, \theta) = \frac{1}{\theta} x^{\frac{1}{\theta} - 1}, \; x \in (0, 1)$ find the distribution of: $Y = - \frac{1}{\theta} \ln X$ [Solution provided: $Y \sim \mathcal{E}(1)$ ]

I did: $P(Y = y) = P(- \frac{1}{\theta} \ln X = y) = P( X = e^{-\theta y}),\; y \in (0, +\infty)$ and so: $f_Y(y, \theta) = f_X(e^{-\theta y}, \theta) = \frac{1}{\theta} e^{-y(1-\theta)}$

Am I missing something?

Is this distrubition a $\mathcal{E}(1)$?If so why?

2 Answers 2

4

If I may quote another answer:

The simplest and surest way to compute the distribution density or probability of a random variable is often to compute the means of functions of this random variable. In the case at hand, one wants to write $\mathrm E(u(Y))$ as $ \color{blue}{\mathrm E(u(Y))=\int u(y)g(y)\mathrm{d}y}, $ for every bounded measurable function $u$. Then one can be sure that $g$ is the density of the distribution of $Y$. So, in a way, the functions $u$ play the role of a dummy variable and one wants the equality above to hold for every $u$.

The rest is easy: introducing $a=1/\theta$ for notational convenience and using the fact that for every bounded measurable function $v$, by definition of the distribution of $X$, $ \mathrm E(v(X))=\int_0^1 v(x)ax^{a-1}\mathrm dx, $ one gets $ \mathrm E(u(Y))=\mathrm E(u(-a\log X))=\int_0^1 u(-a\log x)ax^{a-1}\mathrm dx=\int_0^1 u(-a\log x)x^{a}\frac{a\mathrm dx}x. $ The change of variable $y=-a\log(x)$ yields $y\gt0$, $x^a=\mathrm e^{-y}$ and $\mathrm dy=a\mathrm dx/x$ hence $ \mathrm E(u(Y))=\int_0^{+\infty} u(y)\mathrm e^{-y}\mathrm dy. $ This proves that the density $g$ of $Y$ is defined by $g(y)=\mathrm e^{-y}$ if $y\gt0$ and $g(y)=0$ otherwise. In other words, $Y$ is a standard exponential random variable.

  • 0
    He is obviously writing P{Y=y) meaning the density f(y) and may not realize the difference between a density and a discrete probability mass function. So the problem with his mathematics is that although he made the substitution correctly he left out the Jacobian of the transformation. Isn't that the proper way to answer the question rather than confusing the issue with calculating expectations or belaboring the point about P(Y=y)=0 when$Y$has a probability density function?2012-06-23
2

You can't look at point probabilities when dealing with absolutely continuous random variables. Both $X$ and $Y$ are absolutely continuous (i.e. they have a density wrt. the lebesgue measure) and hence $P(X=x)=P(Y=x)=0$ for all $x\in\mathbb{R}$. Instead try looking at $P(Y\leq y)$ and see if it matches the corresponding CDF of a $\mathcal{E}(1)$ distribution. Edited.

  • 0
    @StefanHansen I made the change.2019-04-21