X is a continuous random variable distributed uniformly over the unit interval [0, 1]. Let Y = 3X + 1. What is the density of Y evaluated at y=4?
$CDF of Y = P(Y \le y) = P(3X + 1 \le y) = P(X \le \frac{y-1}{3})$
$= \int_{-\infty}^\frac{y-1}{3} 1 dx$
Since $0 \le X \le 1$, I reason that the lower limit of integration is 0, i.e. $\int_0^\frac{y-1}{3} 1 dx = \frac{y-1}{3}$
Now to get the PDF, I differentiate to get
$\frac{d}{dy} \frac{y-1}{3} = \frac{1}{3}$
Am I doing the right thing? It doesn't seem right to me because I'm not evaluating at y=4…
EDIT: Never mind, I realized I could check this simply by drawing the distribution of Y, which is uniform over [1, 4]. The PDF integrates to 1, so $\frac{1}{3}$ is the correct answer.
Best Answer
It is correct. You have found the hard way $f_Y(y)=\tfrac 13\mathbf 1_{1\leqslant y\leqslant 4}$ . The required evaluation is easy; clearly $f_Y(4)=\tfrac 13$.
Edit: And yes, $X\sim\mathcal U[0;1]$ so the linear transformation gives: $3X+1\sim\mathcal U[1;4]$