Let X be a random variable uniformly distributed on [0,1] and let Y = [6X] + 1. Determine the distribution of Y and compute E(Y).
My attempt at solving this:
My professor advised us to always consider the values that the function of the RV takes on. As Y takes an integer value, it is a discrete RV. Also, As $X$~Uniform$[0,1]$, Y is at least 1 and at most 7, and is monotone, so can conclude that Y can take values $\in[1,7].$
Now the simplest way that occurs to me to find the distribution (here, the problem is asking for the CDF) is to mechanically compute the PMF of Y. We can do this because if $X\in[0,\frac1{12}]$ (has probability $\frac1{12} - 0 = \frac1{12})$, $Y = 1$. If $X\in[\frac1{12},\frac3{12}]$ (has probability $\frac3{12} - \frac1{12} = \frac1{6})$, $Y = 2$, and so on. I can carry this on to find the values of the PMF of Y, and as Y has 7 values, it shouldn't take too long to do this. Then, once you have the PMF, describing the CDF is simple enough.
My question is: is there a better way to do this? We never formally introduced the 'integer' function, which I understand it from my background in programming as just a rounding function. With other functions of RVs, we would typically find the distribution by doing the following. Consider this example: $X$~$Poisson[1]$, $Y=e^X,$ find the PMF of Y. We start with $F_Y(y) = P(Y<y) = P(e^X < y) = P(X < log(y)) = F_X(log(y)),$ then differentiate to find that $f_Y(y) = \frac1yf_X(log(y))$ for $log(y) \geq 0.$ Is there a way to manipulate the integer function like this, so that in the original problem, you can put the CDF of Y in terms of the CDF of X? Or just a generally better way to approach this problem? (Better meaning more generalizable of an approach)