Determine the Cumulative Distributive Distribution(CDF) of a truncated value?

Question

It is the last part(part h) that I am having problems with.

I know you use integration and then split it into 2 parts.

But how exactly do you do it ? A detailed answer would be very helpful !

Please help !

Answer 1

For $u\lt b$, $F_U(u)=F_Y(u)$.

For $u\ge b$, $F_U(u)=1$. For whenever $Y\ge b$, we have $U=b$. It follows that $U\le b$ always.

That's all there is to it.

Answer 2

It is a lot simpler than you might think. Here is a hint: there are two cases if $U = \min\{Y, b\}$. Either $Y < b$ or $Y \ge b$. If the latter, then $\Pr[U = b] = \Pr[Y \ge b] = S_Y(b)$. Otherwise, $Y < b$ implies $U = Y$ and you already know the density for that situation.

Answer 3

$Pr(U = b) = Pr(Y \geq b) = \int_{b} ^{\infty} f_Y(y) dy$

$f_U(u) = f_Y(u)$ for $u < b$

Hence, the CDF of $u$ will be ,

$F_U(u) = F_Y(u)$ for $u < b$

$F_U(u) = 1 $ for $u \geq b$ as $F_U(b) = F_Y(b) + Pr(U = b) = 1 $, and it will stay at $1$.

The best way to explain this is that $F_U(u_0)$ is the probability that $u \leq u_0$. So, as the variable $u$ is truncated at $b$, so it will be always less than $b$ and any value greater than $b$.