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Let $X\sim \text{Poisson}(\lambda)$ and $Y = \min(X,3)$. Find the pmf for $Y$.

I start out by writing $P(Y=k)=P(\min(X,3)=k)$. If $X\leq 3$, then $Y=X$ and so $Y$ will follow the pmf $f(x)$ for $X$. I am a bit confused about the $X>3$ case. If $X>3$, then $Y = 3$. Therefore, $P(Y=k)=0 \hspace{2mm} \forall k \neq 3$. However, I am not really sure how to tie this into the formulation for the pmf. Would it simply be something like: $$P(Y=k)=\begin{cases}P(X=k) && X\leq 3 \\ 1 && k=3 \text{ and }X>3 \\ 0 && \text{o.w.} \end{cases}$$ Though this looks to me more like $f(y|x)$, in which case I need $f(x,y)$ which I am not sure if I can find it with the given information. Furthermore, I am not really sure how to capture the "$k=3$ and $X>3$" part as this seems rather inelegant to include into a pmf.

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  • $\begingroup$ Probabilities for values $0,1,$ and $2$ remain unchanged. See my Answer for more detail. $\endgroup$
    – BruceET
    Commented Feb 11, 2021 at 7:09

1 Answer 1

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Here is a simulation in R, using $\lambda=5,$ that shows approximate results and histograms. With a million iterations simulated results will usually be accurate to 2 or 3 places--good enough for us to see what's happening.

set.seed(2021)
lam=5
x = rpois(10^6, lam)

We get simulated values of $X \sim\mathsf{Pois}(\lambda=5)$ from $0$ through $20.$ The table below shows approximate probabilities in the PDF of this distribution up to $20.$

summary(x)
       Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      0.000   3.000   5.000   5.002   6.000  20.000 
table(x)/10^6
x
       0        1        2        3        4        5        6 
0.006852 0.033641 0.083564 0.140302 0.176180 0.175577 0.146213 
       7        8        9       10       11       12       13 
0.104059 0.065047 0.036384 0.018425 0.008359 0.003408 0.001281 
      14       15       16       17       18       19       20 
0.000476 0.000146 0.000065 0.000014 0.000003 0.000003 0.000001 

After we take the minimum $Y = \min(X,3),$ all of the probability $P(X \ge 3) = 0.8753$ will be put onto the event $\{Y=3\}.$

1 - ppois(2,lam)
[1] 0.875348

Now we find the approximate distribution of $Y.$ Its PDF is the same as that of $X$ for values $0,1,$ and $2,$ but $P(Y=3) = 0.8753,$ approximated here as 0.875943. The probabilities of values $0,1,2$ remain unchanged.

y = pmin(x,3)
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   3.000   3.000   2.829   3.000   3.000 
table(y)/10^6
y
       0        1        2        3 
0.006852 0.033641 0.083564 0.875943 

Exact values, to four places, are as follows:

pdf.y = c(dpois(0:2, 5), 1-ppois(2, 5))
round(pdf.y, 4)
[1] 0.0067 0.0337 0.0842 0.8753

Here are histograms of random variables $X$ and $Y.$ The area under both histograms to the right of the vertical dashed line is the same.

enter image description here

par(mfrow=c(1,2))
 hist(x, prob=T, br=seq(-.5,20.5), col="skyblue2",
      main="POIS(5)")
  abline(v=2.5, col="red", lwd=2, lty="dotted")
 hist(y, prob=T, br=seq(-.5,3.5), col="skyblue2",
      main="min(POIS(5),3)")
  abline(v=2.5, col="red", lwd=2, lty="dotted")
par(mfrow=c(1,1))
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