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Let $X$ a real random variable with density $f_X$ and $Y=g(X)$ with $g$ measurable.

I have some trouble understanding the following statement :

" If $g(x)=a$ with $a$ a constant, for all $x$, the distribution of $Y$ is the Dirac mass in $a$, which doesn't have a density "

My questions :

1 - How can we translate this statement into formulas?

2 - Why doesn't the density exist?

Dave ddd
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2 Answers2

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Let $(\Omega,\mathcal A,\Pr)$ denote the probability space that serves as domain for random variable $X:\Omega\to \mathbb R$. Then $\{Y\in A\}=\{\omega\in\Omega\mid g(X(\omega))\in A\}=\Omega$ if $a\in A$ and $\{Y\in A\}=\varnothing$ otherwise.

1) Let $P_Y$ denote the probability meausure on $(\mathbb R,\mathcal B)$ prescribed by $A\mapsto\Pr(Y\in A)$.

That leads to:$$P_Y=\delta_a$$

This because $P_Y(A)=\Pr(Y\in A)=1$ if $a\in A$ and $P_Y(A)=\Pr(Y\in A)=0$ otherwise.

2) The distribution (it exists all right) will have function $F(x)=1_{[a,\infty)}(x)$ as cumulative distribution function. In order to have a density function the CDF must at least be continuous, which is not the case.

drhab
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  1. Unpack the variable $Y$: If $c < d$ are real numbers, then $$ P(c < Y < d) = P(c < g(X) < d) = P(c < a < d) $$ This is 1 if $a$ is between $c$ and $d$, and $0$ if it isn't.
    So by extension, the distribution of $Y$ is identical to the distribution of a point mass centered at $a$: $$ \mu_Y(E) = P(Y \subseteq E) = \begin{cases} 1 & a \in E \\ 0 & a \notin E \end{cases} $$

  2. The distribution of $Y$ doesn't have a density function. To see why, suppose it did have a density $f_Y$. let $E$ be the set of numbers $\neq a$. Then for any measurable set $E'\subseteq E$, $$ \int_{E'} f_Y\,dx = \mu_Y(E') = 0 $$ since $a \notin E'$. This means that $f_Y = 0$ almost everywhere in $E$. Replacing $a$ (a set of measure zero by itself), we still have $f_Y = 0$ almost everywhere in $\mathbb{R}$. But $$ \int_{\mathbb{R}} f_Y\,dx = \mu_Y(\mathbb{R}) = 1, $$ a contradiction.

Matthew Leingang
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  • Very clear for the first point. In the second one, How did we conclude that $f_Y=0$ almost everywhere in $E$ implies that we still have that almost everywhere in $R$? thank you – Dave ddd Jul 12 '16 at 16:22
  • @Daveddd: “almost everywhere” implies with respect to Lebesgue measure, and the Lebesgue measure of discrete sets (like singletons) is zero. So if $F$ is the set of points of $\mathbb{R}\setminus\left{a\right}$ on which $f$ is nonzero, and $F'$ is the set of points of $\mathbb{R}$ on which $f$ is nonzero, the difference between $F'$ and $F$ is either the singleton $\left{a\right}$ or nothing at all. This means $F$ and $F'$ have the same Lebesgue measure. We already showed $F$ has measure zero. – Matthew Leingang Jul 12 '16 at 16:35
  • So this is the same as saying that $R \backslash { a }$ and $ { a }$ are disjoint, so $\mu_Y(R)=\mu_Y(R \backslash { a })+\mu_Y( { a })$, and since $\mu_Y( { a }) = 0$ and $\mu_Y(R \backslash { a })=0$ then $\mu_Y( { a })=0$. Thank you – Dave ddd Jul 13 '16 at 14:36
  • @Daveddd: Not exactly. $\mu_Y(\mathbb{R}) = \mu_Y({a}) =1$. The problem that (using $\lambda$ for Lebesgue measure) $\mu_Y({a})=1$ while $\lambda({a}) = 0$. So $\mu_Y$ is not absolutely continuous with respect to Lebesgue measure, which means it can't have a Radon-Nikodym derivative (i.e., density). See this answer to a related question. – Matthew Leingang Jul 13 '16 at 17:41