# Let $X$ be a random variable with probability density function $f$. Prove that the probability distribution function of $X$ is non-decreasing

Recall from calculus that a function $h$ is called non-decreasing if $x \le y$ implies $h(x) \le h(y)$, for every $x, y \in \mathop{\mathrm{dom}} h$.

Q1a) Let $X$ be a continuous random variable with probability density function $f$. Prove that the probability distribution function of $X$ is non-decreasing.

I'm assuming this means show $F(x) = \int_{-\infty}^x f(y)\,dy$, is a non-decreasing function of $x$ in $\mathbb R$.

Q1b) Show that $\lim_{x\to-\infty} F(x) = 0$ and $\lim_{x\to \infty} F(x) = 1$, and explain the probabilistic meaning of these facts.

Sorry about the layout i'm not used to using this site, hope it makes sense!

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Hint for Q1a) $f \ge 0$. –  martini May 10 '12 at 9:48
Hint for Q1b) $F(x) = \lim_{a \to -\infty} \int_{a}^x f(y)\ dy = \lim_{a \to -\infty} (F(x) - F(a))$. –  Robert Israel May 11 '12 at 1:39

One way to do 1(b): $F(x) = \int_{-\infty}^x f(t)\ dt$ is an improper integral, which by definition of improper integral means $\lim_{a \to -\infty} \int_a^x f(t)\ dt$. Now $\int_a^x f(t)\ dt = F(x) - F(a)$, so $$F(x) = \lim_{a \to -\infty} (F(x) - F(a)) = F(x) - \lim_{a \to -\infty} F(a)$$ and you can solve for $\lim_{a \to -\infty} F(a)$.

As for $\lim_{x \to \infty} F(x) = \lim_{x \to \infty} \int_{-\infty}^x f(t)\ dt$, that is $\int_{-\infty}^\infty f(t)\ dt$, which according to the definition of a probability density function must be $1$.

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The proof by Robert on the upper limit does depend on f approaching ∞ as x does. Then to know that the limit equals 1 relies on the definition of a probability density. That f tends to 0 as x approaches +∞ or −∞ is a required property for a probability desnity function. –  Michael Chernick May 13 '12 at 15:29
@MichaelChernick: where do you see $f(x) \to 0$ as $x \to \infty$ written as a required property for a probability density? The only real requirements are $f \ge 0$ and $\int_{-\infty}^\infty f(x)\ dx = 1$. Some elementary texts (wanting to avoid Lebesgue integration) may require $f$ to be piecewise continuous. But I haven't seen any that required limits of $0$. –  Robert Israel May 13 '12 at 17:59
As I said it is required because it would be impossible for ∫f(x) dx=1 integrating over (-∞, ∞) if f does not go to 0 as x approaches ∞!!! –  Michael Chernick May 13 '12 at 20:52
No. For example, try an $f$ whose graph consists of infinitely many triangular "bumps", the $n$'th bump (for positive integers $n$) having height $2n$ and width $2^{-n}/n$, so area $2^{-n}$. –  Robert Israel May 13 '12 at 21:17
Or if you prefer an analytic function, try $6 \pi^{-5/2}\sum_{n=1}^\infty n e^{-n^6 (x-n)^2}$. –  Robert Israel May 13 '12 at 21:24
Not at all silly. Actually $f$ doesn't have to go to $0$ as $x \to + \infty$ or $-\infty$. Fortunately, that's irrelevant to the questions being asked. –  Robert Israel May 11 '12 at 1:36