# A random variable has a PDF that is proportional to x on the interval [0,1] and zero elsewhere. What is the formula for this pdf?

A random variable has a Probability density function that is proportional to x on the interval [0,1] and zero elsewhere. What is the formula for this pdf? What is the cdf?

So I understand that the formula is essentially the equation reflecting the curve of the function. I'm a bit confused by the terminology "proportional to x."

"Proportional to $x$" means that it is of the form $f(x)=cx$ for some constant $c$.
Now, for it to be a probability density, it needs to integrate to 1, so $$\int_\mathbb Rf(x)\,dx = \int_0^1 cx\,dx = c \left[\frac{x^2}{2}\right]_0^1 = c\left(\frac{1}{2}-\frac{0}{2}\right)=\frac{c}{2}=1.$$
From this you can figure out the value of $c$ that makes the function be a probability density.
EDIT: once you have the density function, you can also find the cumulative distribution function by integration: for every $x$ in $[0,1]$, $$F(x)=\int_{-\infty}^x f(t)\,dt=\int_0^x 2t\,dt = 2\left[\frac{t^2}{2}\right]_0^x=(x^2-0)=x^2.$$
You can check it fulfills all the conditions to be a cumulative distribution: it is always nonnegative, when $x\rightarrow -\infty$ it tends to $0$ (because $F(0)=0$), when $x\rightarrow +\infty$ it tends to $1$ (because $F(1)=1$), and it is right-continuous (it is actually continuous).
• hmmm. How did you get $$c \left[\frac{x^2}{2}\right]_0^1$$ from the question? – aero26 Mar 14 '17 at 2:21
• $G(x)=\frac{x^2}{2}$ is the primitive of $g(x)=x$, and then applied the fundamental theorem of calculus. Maybe the notation of square brackets is not that standard? – Anna SdTC Mar 14 '17 at 2:25