I am trying to unite my knowledge of statistics and measure theory by considering the following example.

Suppose we have a measurable space $(\Omega_1,B_1)$ and a random variable (measurable function) on the space, call it $X$: $\Omega_1 \rightarrow R$.

Suppose we know the distribution function of $X$, say it is normal $X \sim N(0,1)$. Now consider the random variable $$Y=X +5$$

We know from basic statistics that $Y\sim (5,1)$, but how can we prove that by using the definition of $X$ and composition of functions? Explanations that are step by step greatly appreciated!


In order to properly speak of a random variable's distribution, you need a measure space (with a measure), and specifically a probability space $(\Omega, \mathscr A, \mathbb P)$, where of course $\mathbb P (\Omega) = 1$.

The distribution of a random variable $X$ is the image measure $X(\mathbb P)$, i.e.

$$ X(\mathbb P)(A) :=\mathbb P(X^{-1}(A)) \quad \text{for }A\in\mathscr A,$$

Outside of the measure-theory context, I first encountered this as the 'cumulative distribution function' of a random variable, i.e. the function that gives the probability of $X$ being less than or equal to $x$:

$$ F_X(x) = \mathbb P (X \leq x) = \mathbb P \left(X^{-1} (-\infty, x]\right).$$

You may recognise $(-\infty, x]$ as being the sets that generate the Borel sets on $\mathbb R$.

A random variable has normal distribution $N(0,1)$ when this image measure is given by:

$$ X(\mathbb P)(A) = \int_A e^{-x^2/2} \, \mathrm \lambda(\mathrm dx),$$

where $\lambda$ is the one dimensional Lebesgue-measure. Phrased in less measure-theory heavy terms,

$$ F_X(x) = \int_{-\infty}^x e^{t^2/2} \, \mathrm dt.$$

If $Y = X+5$, then we have $Y(\mathbb P)(A)=X(\mathbb P)(A-5)$, giving

$$ Y(\mathbb P)(A) = \int_{A-5} e^{x^2/2} \, \lambda(\mathrm dx)\\ =\int_A e^{(x-5)^2/2} \, \lambda(\mathrm dx),$$

which is the definition of $Y$ having $N(5,1)$ distribution. Here we used the fact that $x \mapsto x +5$ is measurable (because it is continuous) to justify the last step.


As S. van Nigtevecht says, when you say $X\sim N(0,1)$ you are saying there is a probability measure added to the setup, so we have in all a measure space $(\Omega, B, P)$, and a pair of functions $X$ and $Y$ such that $Y(\omega)=X(\omega)+5$ for all $\omega\in\Omega$. If you want, $Y$ is the composition of $X$ with the "add to 5" function. We also know also that $P(\{\omega: X(\omega)\le a\})=\int_{-\infty}^a \phi(t)\,dt$, where $\phi$ is the $N(0,1)$ density function. So $Y^{-1}([-\infty,x]) = X^{-1}([-\infty,x+5]),$ and $$P(Y\le x) = P(X\le x+5 ) = P(\{\omega: X(\omega)\le x+5\}) = \int_{-\infty}^{x+5}\phi(t)\,dt$$ and so on. (I have dropped the subscript $1$ here & there.)

Is this what you want?


As other have mentioned, you first need a probability measure on $\Omega_1$. The statement $Z \sim N(\mu,\sigma^2)$ can then be defined as saying that for any bounded measurable function $\phi: \Bbb R \to \Bbb R$, we have that $$E[\phi(Z)] = (2\pi \sigma)^{-1/2}\int_{\Bbb R} \phi(z)e^{-(z-\mu)^2/(2\sigma^2)}dz$$ or in simpler terms, the density of $Z$ (more technically the Radon-Nikodym derivative of the law of $Z$ with respect to Lebesgue measure) is a multiple of $e^{-(z-\mu)^2/(2\sigma^2)}$.

Now let $X \sim N(0,1)$ and $Y:=X+5$. Given any bounded measurable $\phi$ we define $\psi(x):=\phi(x+5)$ and we find that $$E[\phi(Y)] = E[\psi(X)] = c\int_{\Bbb R} \psi(x)e^{-x^2/2}dx = c\int_{\Bbb R} \phi(x+5)e^{-x^2/2}dx=c\int_{\Bbb R} \phi(y)e^{-(y-5)^2/2}dy$$ where $c= (2\pi)^{-1/2}$. This shows that the density of $Y$ is $e^{-(y-5)^2/2}$, or in other words, $Y \sim N(5,1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.