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Can anyone please help me with this random variable question I've stumbled across.

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 Q1b) $F(x) = \lim_{a \to -\infty} \int_{a}^x f(y)\ dy = \lim_{a \to -\infty} (F(x) - F(a))$.2012-05-11

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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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    Maybe between mathematicians and statisticians. Maybe this illustrates a difference between statisticians and mathematicians. We think of distributions with nice smooth densities all of which go to$0$in the right and left tails while matematicians think of pathological cases to show counterexamples to our ideal.2012-05-14
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Yes because the density function is never less than 0 the integral from -∞ to x cannot be larger than from -∞ to y for any y>x. So F is monotone non-decreasing. The density function f has to go to 0 as x approaches + or - ∞. So lim x→−∞ F(x)=0. The fact that lim x→∞ F(x)=1 is a requirement for f to be a probability density with F as its CDF. There is no other way to see that f integrates to 1 from the information given. All you can deduce is that it integrates to a positive constant.

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    @MichaelChernick Start with a 'nice' density $f$ that goes to $0$ if $x$ tends to $+\infty$. Then define $g$ by $x\mapsto5$ if $x\in\mathbb{Q}$ and $x\mapsto f\left(x\right)$ otherwise. This less nice $g$ can - just like $f$ - also serve as density, and this for the same distribution. However, it does not go to $0$ if $x$ tends to $+\infty$. Robert is right in this.2013-10-17