PDF for $\xi + \eta$

Question

Let $\xi $ and $ \eta$ be independent random variables, $\xi$ is $U[0,1]$ , $\eta$ takes values $1/3$ and $2/3$ with probability $1/2$. Find the PDF for $\xi + \eta$
It is easy for two continuous random variables, but what should I do in this situation?

Answer 1

Preliminary Lemma: if $f$ is a pdf, $$f(x)*\delta_a(x)=f(x-a).$$

Proof: Let $X$ a random variable with pdf $f$ and $Y=a$ (constant), thus with pdf $\delta_a$, ("a probability peak of value 1 concentrated in position $a$"). Then the pdf of random variable $X+Y=X+a$, which is clearly the shifted version: $x\mapsto f(x-a)$, is besides, the pdf of a sum of random variables, thus the convolution of the associated pdfs.

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The pdf of Random Variable $X$ is $\Pi_{[0,1]}$, the characteristic function of interval $[0,1]$.

The pdf of Random Variable $Y$ is $\frac12(\delta_{1/3}+\delta_{2/3})$.

The pdf of $X+Y$ is the convolution:

$$g=\Pi_{[0,1]} * \frac12(\delta_{1/3}+\delta_{2/3})$$

By distributivity property:

$$g=\frac12\Pi_{[0,1]} * \delta_{1/3}+\frac12\Pi_{[0,1]} * \delta_{2/3}$$

Using the Lemma, one can write:

$$g=\frac12\Pi_{[1/3,4/3]}+\frac12\Pi_{[2/3,5/3]}$$

which can be also written:

$$g=\frac12\Pi_{[1/3,2/3]}+\Pi_{[2/3,4/3]}+\frac12\Pi_{[4/3,5/3]}$$

The graphical representation of this pdf is:

Answer 2

Comment: Here is a simulation in R statistical software that may help you verify the answer when you get it. I used $X \sim Unif(0,1)$ and $Y$ taking values $1/3$ and $2/3$ each with probability $1/2,$ and $S = X + Y.$ With a million realizations of each random variable simulated means and SDs should be accurate to about three places.

m = 10^6;  x = runif(m)
y = sample(c(1/3,2/3), m, repl=T)
s = x + y
cut = seq(1/3, 5/3, 1/18)
hist(s, prob=T, br=cut, col="wheat")
mean(x); mean(y);  mean(s)
## 0.4995831      # aprx E(X) = 1/2
## 0.500128       # aprx E(Y) = 1/2
$$ 0.9997111      # aprx E(S) = 1
sd(x); sd(y); sd(s)
## 0.28881
## 0.1666667
## 0.3337064
summary(s)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.3333  0.7492  0.9993  0.9997  1.2500  1.6670
quantile(s, (1:6)/6)
## 16.66667% 33.33333%       50% 66.66667% 83.33333%      100% 
## 0.6659555 0.8329647 0.9993092 1.1662825 1.3340041 1.6666663 

You can view this as a 50:50 mixture of two uniform distributions, one of them on the interval $(1/3, 4/3).$