Random variables defined on the same probability space with different distributions

Consider the real-valued random variable $X$ and suppose it is defined on the probability space $(\Omega, \mathcal{A}, \mathbb{P})$. Assume that $X \sim N(\mu, \sigma^2)$. This means that $$(1)\text{ } \mathbb{P}(X\in [a,b])=\mathbb{P}(\{w \in \Omega \text{ s.t } X(\omega)\in [a,b]\})=\frac{1}{2}\left(1+\frac{1}{\sqrt{\pi}}\int_{-(\frac{x-\mu}{\sigma \sqrt{2}})}^{(\frac{x-\mu}{\sigma \sqrt{2}})}e^{-t^2}dt\right)$$ In several books I found that we can also say that $X$ is distributed according to $\mathbb{P}$.

Now suppose that we add another random variable $Y$ on the same probability space and assume $Y \sim U([0,1])$. This means that, for $0\leq a\leq b \leq 1$ $$(2)\text{ } \mathbb{P}(Y \in [a,b])=\mathbb{P}(\{w \in \Omega \text{ s.t } Y(\omega)\in [a,b]\})=b-a$$

Question: the fact that $X$ and $Y$ are defined on the same probability space but have different probability distribution is a contradiction? What is the relation between $\mathbb{P}$, the normal cdf and the uniform cdf? Can we say that both $X$ and $Y$ are distributed according to $\mathbb{P}$ even if they have different distributions?

• I am confused by the statement that $X$ is distributed according to $\mathbb{P}$. $\mathbb{P}$ is a probability measure, and the distribution function $F$, which can be Normal, Uniform, Exponential..., are all induced by that measure. That is $F(x) = \mathbb{P}((-\infty, x))$ and it is the distribution function that characterizes the probability measure. So we may determine from $F$ the probability $\mathbb{P}(A)$ for any Borel set A. – Bob Dobalina Dec 16 '15 at 14:38
• Ok, thanks a lot. Could you expand in an answer clarifying the statement "it is the distribution function that characterizes the probability measure. So we may determine from $F$ the probability $P(A)$ for any Borel set $A4" when we have different random variables with different cdf define don the same probability space? – TEX Dec 16 '15 at 14:48 3 Answers Admittedly, a holistic answer to your questions would require more measure-theoretic machinery than what follows. However, I will attempt to give you succinct responses that you might find helpful. So, let the real-valued random variables$X, Y$be defined on the same probability space$(\Omega, \Sigma, \mathbb P)$. 1)$X$and$Y$are measurable so that, for instance, for the interval of real numbers$[a,b]$, we necessarily have$\left\{X\in[a,b]\right\}, \left\{Y\in[a,b]\right\} \in \Sigma$, while we need not have $$\{X\in[a,b]\} = \{Y\in[a,b]\}.$$ 2) Because of 1) above, we need not have $$\mathbb P\left\{X\in[a,b]\right\} = \mathbb P\left\{Y\in[a,b]\right\}.$$ 3) Note that, because we may define the probability measure$\mathbb P_X(B):=\mathbb P\{X \in B\}$over Borel sets$B \in \mathcal B(\mathbb R)$, we can speak of$X$being distributed according to$\mathbb P_X$. In so doing, we are thinking of$X$in terms of the probability space$(\mathbb R, \mathcal B(\mathbb R), \mathbb P_X)$, not the probability space$(\Omega, \Sigma, \mathbb P)$. In your example, since$X\sim N(\mu, \sigma^2)$, we have an integral representation of$\mathbb P_X$with respect to the Lebesgue-measure, so that $$\mathbb P_X([a,b])=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}{\bf{1}}_{[a,b]}(x)e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\mathrm dx\,.$$ A similar development holds for the uniform random variable$Y$. 4) All of the foregoing is just one way of proceeding; there are alternatives. For instance, one may define$X, Y$over the same measurable-space$(\Omega, \Sigma)$, but different probability-spaces,$(\Omega, \Sigma, \mathbb P_X)$and$(\Omega, \Sigma, \mathbb P_Y)$, with different probability measures. First of all, we have two spaces: the probability space$(\Omega,\mathcal A,\mathbb P)$and the measurable space$(\mathbb R,\mathcal B(\mathbb R))$, where$\mathcal B(\mathbb R)$is the Borel$\sigma$-algebra of$\mathbb R$, i.e. the smallest$\sigma$-algebra that contains all open sets. A random variable$X$is a measurable function that maps$\Omega$to$\mathbb R$. Measurable means that$X^{-1}(B)=\{\omega\in\Omega:X(\omega)\in B\}\in\mathcal A$for each$B\in\mathcal B$. Rouhgly speaking, randomness takes place in the probability space. So we can calculate the probability of an event$A\in\mathcal A$. It is given by$\mathbb P(A)$. However, we are interested in the events$B\in\mathcal B(\mathbb R)$. Measurability enables us to evaluate the probabilities of such events. We have that$\mathbb P(B)=\mathbb P\{\omega\in\Omega:X(\omega)\in B\}$and here we have the random variable$X$in the expression. If we take another random variable$Y$, the probability of the event$B$is then given by$\mathbb P(B)=\mathbb P\{\omega\in\Omega:Y(\omega)\in B\}$and these probabilities might be different. In general, these probabilities depend on two objects: a random variable and the probability measure$\mathbb P$. The distribution of the random variable$X$is the probability measure defined on$\mathcal B(\mathbb R)$by setting$\mathbb P(B)=\mathbb P\{\omega\in\Omega:X(\omega)\in B\}$and this distribution might be the uniform distribution, the normal distribution or any other probability measure on$\mathcal B(\mathbb R)$. So the fact that$X$and$Y$are defined on the same probability space, but have different probability distribution is not a contradiction. The distribution depends on the random variable, so if we take another random variable defined on the same probability space, we obtain a different distribution. I hope this helps. TL;DR I think the source of your confusion is seeing$X$and$Y$as being both the identity random variable in$(\Omega, \mathscr{F},\mathbb P)$. I'm going to give an example of explicit exponential and uniform distributions in the same probability space in$(\Omega, \mathscr{F},\mathbb P)$. Consider a random variable$X$in$((0,1), \mathcal{B}(0,1), Leb)$given by $$X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega}, \lambda > 0$$ It has cdf$F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{(0,\infty)}$, which we know to be the cdf of an exponentially distributed random variable. (*) Actually, $$X(1-\omega):=\frac{1}{\lambda} \ln \frac{1}{\omega}, \lambda > 0$$ also has cdf$F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{(0,\infty)}$. Are all cdfs in this mysterious probability space exponential? No! Now consider the identity random variable$U$in$((0,1), \mathcal{B}(0,1), Leb)$: $$U(\omega):=\omega$$ It has cdf$F_U(u) = P(U \le u) = u1_{(0,1)}+1_{(1,\infty)}$, which we know to be the cdf of a uniformly distributed random variable. The above$X$and$U$have different CDFs under the same probability space. The aforementioned explicit representations of the exponential and uniform distributions in this probability space are called Skorokhod representations in$((0,1), \mathcal{B}(0,1), Leb)$. Now consider the identity random variable$U$in$(\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$No surprise that$U$is exponential by definition:$F_U(u) = P(U \le u) = (1-e^{-\lambda \{u\}})1_{(0,\infty)}$. Now you're wondering: Aha! So every random variable here is exponential right? Well no, for any distribution you can think of, say, uniform, Bernoulli, etc, all have a place here and their Skorokhod representaion is given by: $$Y(\omega) = \sup(y \in \mathbb{R}: F(y) < \omega)$$ Try for yourself to see for yourself that $$Y(\omega) = \sup(y \in \mathbb{R}: y1_{(0,1)}+1_{(1,\infty)} < \omega)$$ has uniform distribution in$(\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$, i.e. $$P(Y \le y) := P(\sup(y \in \mathbb{R}: y1_{(0,1)}+1_{(1,\infty)} < \omega) \le y) = y1_{(0,1)}+1_{(1,\infty)}$$ Also try to see for yourself that$X(\omega)$above no longer has exponential distribution in this probability space. (**) Conclusion: I think the source of your confusion is seeing$X$and$Y$as being both the identity random variable in$(\Omega, \mathscr{F},\mathbb P)$. If you were to see them explicitly, you would know that they definitely don't necessarily have the same distribution. What$\mathbb P$does is tell you the probabilities of$\omega$'s. So you know how likely the sample point $$0.5 \in \Omega = (0,1)$$ is but not directly how likely the random variable$X$is equal to a number in its domain such as the real number $$X(0.5) = \frac{1}{\lambda} \ln \frac{1}{1-0.5} \in \mathbb R$$ is. (*) Of course, the probability that$X$is the real number$X(0.5)$is 1. dependent on the probability that the sample point$0.5$because$0.5=1-e^{-\lambda X(0.5)}$2. not expected to be the same in another probability space, assuming of course that$X$is in the new probability space, because it now depends on the probability of the sample point/s$X^{-1}(X(0.5))$aka$X \in \{X(0.5)\}$. (**) Pf of (*): Two steps in computing$P(X \le x)$: 1. Find all$\omega \in \Omega = (0,1)$s.t.$X(\omega) \le x$2. Compute the probability of all those$\omega$'s. For$x \le 0$,$P(X\leq x) = P(X \in \emptyset^{\mathbb R}) = P(\emptyset^{\Omega}) = 0$For$x > 0$,$X(\omega) \le x$Step 1: $$\iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$ $$\iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$ $$\iff \omega \in (0,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$ $$\iff \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$ Step 2: $$Leb(\omega | \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$ $$= Leb((0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$ $$= \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$ QED Pf of (**): Actually$X \notin (\mathbb R, \mathscr B(\mathbb R), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$because we need$\frac{1}{1-\omega} > 0 \iff \omega < 1$. QED Same for$X(1-\omega)$where we need$\frac{1}{\omega} > 0 \iff \omega > 0$. But we can can further try to show$X$is not exponential in$((-\infty,1), \mathscr B((-\infty,1)), (1-e^{-\lambda \{u\}})1_{(0,\infty)})$Pf: For$x \le 0$,$P(X\leq x) = P(X \in \emptyset^{\mathbb R}) = P(\emptyset^{\Omega}) = 0$For$x > 0$,$X(\omega) \le x\$

Step 1:

$$\iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$

$$\iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$

$$\iff \omega \in (-\infty,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$

$$\iff \omega \in (-\infty,\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}}\})$$

Step 2:

$$P(\omega | X(\omega) \le x)$$

$$= P(\omega | \omega \in (-\infty,\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}}\}))$$

$$= \int_{-\infty}^{\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}})\} d((1-e^{-\lambda \{u\}})1_{(0,\infty)})$$

$$\int_{-\infty}^{\min\{1,\frac{e^{\lambda x} - 1}{e^{\lambda x}})\} \lambda e^{-\lambda u}$$

$$1-e^{-\lambda \min\{1,1-e^{-\lambda t}\}}$$

Doesn't look exponential to me.

QED