The cumulative distribution function (cdf) gives the probability that the random variable $X$ is less than or equal to $x$ and is usually denoted $F(x)$. The cumulative distribution function of a random variable $X$ is the function given by \[F(x)= \mathrm{P}[X \leq x].\]
Let $X$ be a random variable with cdf $F(x)$. Then \[P[a<X \leq b] = F(b) - F(a).\]
Let $X$ be a continuous random variable with pdf $f(x)$. Then the cumulative distribution function $F(x)$ can be expressed by \[F(x) =\int\limits_{- \infty}^x f(t) \mathrm{d} t.\]
The cumulative distribution function $F(x)$ lies within $0$ and $1$ for all values of $x$. In addition to this \[\lim\limits_{x \rightarrow -\infty}F(x) = 0\] and \[\lim\limits_{x \rightarrow \infty}F(x) = 1.\]
Observe, for $a
The pdf of a continuous random variable $X$ is \[f(x) = \begin{cases} 0 & \text{ if } x<1, \\[6 pt] \dfrac{2}{x^3} &\text{ otherwise}. \end{cases}\]
Find the cumulative distribution function $F(x)$.
We need to split this up into two parts, where $x<1$ and where $x \geq 1$. Substitute into the formula provided in the definition for each part.
If $x<1$ we have \[\int\limits_{-\infty}^{x}f(t) \mathrm{d} t =0\]
and if $x\geq 1$ we have \begin{align} \displaystyle\int\limits_{-\infty}^{x}f(t)\mathrm{d} t &= \int\limits_{1}^{x}f(t) \mathrm{d} t \\ & = \int\limits_{1}^{x} \frac{2}{t^3}\mathrm{d} t \\ & = \left[ - \frac{1}{t^2} \right]_{1}^{x} \\ & = -\frac{1}{x^2} + 1\text{.} \end{align}
So we have obtained \[F(x) = \begin{cases} 0 & \text{ if } x<1, \\[6pt] 1-\frac{1}{x^2} &\text{otherwise} . \end{cases}\]
Note that $\lim\limits_{x \rightarrow -\infty} F(x) = 0$ and $\lim\limits_{x \rightarrow \infty} F(x) =1$ as required.
In this video, Daniel Organisciak finds the cumulative distribution of a probability density function.
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