Problem : Let $X$ be r.v following a uniform law on $[0,1]$, find the cdf of $Y=\min(X,a)$ $(a \in [0,1])$ and prove that the law of Y is linear combination of of a law with a density and a Dirac measure.
Solution:
$\forall t < 0, P(Y\leq t) = 0$
$\forall 0\leq t<a ,$ $P(Y \leq t)=P(X \leq t)=t$
$\forall t \geq a, P(Y\leq t) = 1$
The cumulative distribution function admits a discontinuity point in $a$ where it "jumps" from the value $1-a$ to the value 1.
Let $P_Y$ be the law of Y , ($P_Y$ then has a measure of probability on R). We want to write $P_Y$ as a linear combination of a measure of a density (when the measure represents the integral of a density) and of a Dirac.
We notice that $P_Y(]-\infty , t]) = P(Y \leq t)$ by definition, we verify that $P(Y\leq t) = aF_U(t) + (1-a)F_V(t)$ with $F(t)=0$ if $t \leq 0$, $\dots (1) $
$F_U(t) = t/a$ if $t \in [0,a]$
et $F_U(t)=1$ if $t \geq a $
while $F_V(t) = 0$ if $t < a$ and $F_V(t)=1$ if $t \geq a$
We see that $F_U$ is the cumulative distribution function of a uniform law on $[0,a]$ $\dots (2) $
And $F_V$ is a cumulative distribution function of Dirac measure on $a$
We rewrite $ P_Y(]-\infty,t])=aP_U(]-\infty,t]) + (1-a)P_V(]-\infty,t])$ $\dots (3) $
Where $P_U$ is the uniform measure on $[0,a]$ and $P_V$ is that of Dirac in a. $\dots (4) $
The law of $Y$, $(P_Y)$ is then a linear combination of Dirac in $a$ and the uniform law on $[0,a]$ , weighted respectively with $(1-a)$ and $a$
Now, by definition of a Dirac measure, it's a measure supported by a singleton of unitary mass $\forall A \in \mathcal{A}$, and it is a probability measure.
$\mathbb{1}_A(a) =\delta_a(A)= \begin{cases} 1, & \text{if }a \in A \\ 0, & \text{otherwise} \end{cases}$
In this problem, it asks for law of $Y=min(X,a)$ where $X$ is a r.v and $a$ is a fixed given real number in $[0,1]$ in the (1) of the solution , it directly tries to verify a supposed result without giving it's construction , or at least I can't see how to directly come up with that to then just verify it?
for me when I tried to solve it, I assumed a fixed number in its interval and I wrote :
$F_Y(t) = P(min(X,a) \leq t)$
$F_Y = P(\{x \leq t\}\cup\{a \leq t\})$
$F_Y = P(\{x \leq t\}) + P(\{a \leq t\}) -P(\{x \leq t\}\cap\{a \leq t\})$
since I couldn't see something useful coming out of this, I tried another formulation :
$F_Y = P(\{x \leq t\} \cap \{X \leq a\} \cup \{x \geq a\}\cap\{a \leq t\}) $
since a random variable and a constant are independant
$F_Y = at + (1-F(a^{-}))F(\{a \leq t\}) - t $
$F_Y = at + (1-(a^{-})F(\{a \leq t\}) - t $
since the intersection of all the events implies $X=a$ and $X \leq t$ so the probability of that is $t$ hence the $-t$
But then I didn't know what to do with the $F(\{a \leq t\}) $, or use the Dirac measure, what am I not understanding?
EDIT : my reformulation is false particularly for the variables less than $a$ , here's a modified one :
$F_Y = P(\{X \leq t\} \cap \{X \leq a\} \cup \{X \leq t\}\cap\{X \geq a\} \cup \{X \geq t\}\cap\{a \leq t\} ) $
Which then gives :
$F_Y = at + t(1-a^{-})+ F(\{a \leq t\})(1 - t) $ - the intersection of the events two by two + the intersection of all events which I am guessing it's not an efficient way to think about it even if it's correct
A more enlightening solution.
Let $A\in \mathcal B(\mathbb R)$. $$\begin{aligned}P_Y(A) &= P(\min(X,a)\in A) \\ &= E(1_A(\min(X,a)) \\ &= E(1_A(\min(X,a)1_{X\leq a}) + E(1_A(\min(X,a)1_{X> a})\\ &= E(1_A(X)1_{X\leq a}) + E(1_A(a)1_{X>a}) \\ &= E(1_{A\cap[0,a] }(X)) + P(X>a)\delta_a(A) \\ &= P_X(A\cap[0,a])+ P(X>a)\delta_a(A)\end{aligned}$$
Since $P_X(A)=0\implies P_X(A\cap[0,a])=0$, the measure $A\mapsto P_X(A\cap[0,a])$ is absolutely continuous w.r.t $P_X$, hence absolutely continuous w.r.t Lebesgue measure (in other words it has a density).
$A\mapsto\delta_a(A)$ is a Dirac measure with mass $1$.