Probability function of $Y=max\{X,m\}$ for $m$ positive integer when $X$ is geometric distribution

Question

$X$ has geometric distribution so $f_X(x)=p(1-p)^x$. I wrote this: $$f_Y(y)=P(Y=y)=P(max\{X,m\}=y)$$ So if $y=m$ it means that $X<m$ so $f_Y(y)=P(X<m)=1-P(X\ge m)=1-(1-p)^m$ and if $y = m+1,m+2,...$ then $f_Y(y)=p(1-p)^y$

Is it right?

Answer 1

Assuming as correct your $f_X(x)$, that is $X$ is a geometric rv counting the failures before the first success, your pmf $f_Y(y)$ does not sum up to 1...

Observe that $Y=m$ when $X\le m$ and this happens with probability $1-(1-p)^{m+1}$ thus

$$\mathbb{P}[Y=y] = \begin{cases} 1-(1-p)^{m+1}, & \text{if $y=m$ } \\ p(1-p)^y, & \text{if $y=m+1,m+2,m+3,\dots$ }\\ 0, & \text{elsewhere } \end{cases}$$

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As you can see, this pmf works, being

$$1-(1-p)^{m+1}+p\sum_{y=m+1}^{\infty}(1-p)^y=1-(1-p)^{m+1}+p\frac{(1-p)^{m+1}}{1-(1-p)}=1$$