CDf of $Z=\max(a,X)$ when $a$ is a positive constant

Question

If $Z=\max(X,Y)$ where $X$ and $Y$ are continuous independent exponential distributions, the CDF of $Z$ can be written as a product of CDFs of $X$ and $Y$.

However, I need to find the CDF of $Z=\max(a,X)$ where $a$ is a positive constant and $X$ follows an exponential distribution as $F_X(x)=b e^{-b x}$.

can someone please guide me finding this?

Answer 1

Clearly $\mathsf P(Z\leq z) ~{=\mathsf P(\max\{a,X\}\leq z) \\= \mathsf P(a\leq z, X\leq z) \\ = \begin{cases} \mathsf P(X\leq z) &:& a\leq z \\ 0&:& \text{otherwise}\end{cases} }$

So our CDF is $F_Z(z) = (1-\mathsf e^{-bz})\,\mathbf 1_{z\in[a;\infty)}$

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Note: The CDF is $F_X(x)= (1-\mathsf e^{-bx})\mathbf 1_{x\in[0;\infty)}$ if $X\sim\mathcal {Exp}(b)$, while the pdf is $f_X(x)=b\mathsf e^{-bx}\mathbf 1_{x\in[0;\infty}$

Answer 2

First note that if $x < a$ then clearly $P(Z \leq x) = 0$ since $Z$ cannot be less than $a$.

Next, note that if $x = a$ then clearly we have:

$\begin{align} F_Z(x) &= P(Z \leq x)\\ &= P(Z = a)\\ &= P(X \leq a)\\ &= 1 - e^{-ba}\\ \end{align}$

And finally note that if $x > a$ then the probability that $Z < x$ is simply the probability that $X < x$, since $Z < x$ if and only if $X < x$.

As a result we get that:

$F_Z(x) = \begin{cases} \mathsf 1-e^{-bx} &:& a\leq x \\ 0&:& \text{otherwise}\end{cases}$