I need to proof that $X$ follows a distribution $F$ with probability $1-p$ and a distribution $G$ with probability $p$ if, and only if, its distribution function is:

$(1-p)F + pG$

Can anyone give me some hints about how to proof this?

EDIT2: this is my second take based on comments:

$$F_p(x)=Pr(X\le x)=\int_{-\infty}^x dF_p(x)=(1-p)\int_{-\infty}^xdF(x)+p \int_{-\infty}^xdG(x) = (1-p)F+pG$$

but I am not sure if this is a valid proof

  • 1
    $\begingroup$ Are you familiar with the law of total probability? In general, it would be convenient if you could tell us what your level of knowledge is, as there are a number of ways one could go about this. $\endgroup$
    – Stijn
    Jul 16, 2013 at 12:27
  • $\begingroup$ @Stijn, thanks for your comment. Yes I am familiar with basic statistical courses (descriptive statistics, inference, probability theory, etc) $\endgroup$
    – sets
    Jul 16, 2013 at 13:26
  • $\begingroup$ Hint: When $X\sim F$ with probability $1-p$ and otherwise $X\sim G$, you can use your familiarity with the basics to work out $\Pr(X\le x)$ for any $x$. What does it equal? $\endgroup$
    – whuber
    Jul 16, 2013 at 14:20
  • 1
    $\begingroup$ @whuber: starting with $Pr(X\le x)$ i get: $$F_p(x)=Pr(X\le x)=\int_{-\infty}^x dF_p(x)=\int_{-\infty}^xd[(1-p)F+pG](x)=(1-p)\int_{-\infty}^xdF(x)+p \int_{-\infty}^xdG(x) = (1-p)F+pG$$But is that a valid proof? No charactersitic function needed? $\endgroup$
    – sets
    Jul 17, 2013 at 7:59
  • $\begingroup$ No, that is not a valid proof because it is circular: the third equals sign assumes the result! You need to compute $F_p(x)$ using the probabilistic definition of the mixture given in the first line of your question. This can be done in a small number of completely elementary steps (using basic definitions); no characteristic functions are needed. $\endgroup$
    – whuber
    Jul 17, 2013 at 13:17

1 Answer 1


When we state that

$X$ follows a distribution $F$ with probability $1−p$ and a distribution $G$ with probability $p$,

we have implicitly introduced an independent Bernoulli$(p)$ random variable $U$. By definition, this has the value $0$ with probability $1-p$ and the value $1$ with probability $p$. It is used to select between $F$ and $G$. In order to write about this, let $Y$ have distribution $F$ and $Z$ have distribution $G$. All three variables $U, Y, Z$ are independent. Using this notation and the axiom of conditional probability, we can express the chance that $X$ is less than or equal to some arbitrary real number $x$ as

$$\eqalign{ \Pr(X\le x) &= \Pr(U=0 \text{ and } Y\le x) + \Pr(U=1 \text{ and } Z\le x) \\ &= \Pr(U=0)\Pr(Y\le x) + \Pr(U=1)\Pr(Z\le x). }$$

The independence assumptions about $(U,Y)$ and $(U,Z)$ assure the second equality. The result follows immediately from the definitions of $F$ and $G$.


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