# Finding pmf of random variable with minimum function

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.

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

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.

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))