# Expected value of the minimum of a non-negative random variable and a constant

X is a non-negative random variable. Define Y = MIN(X, c) where c is a constant. What is E[Y]? I am modeling the constant as another random variable whose pdf is Dirac Delta function: $f_{c}(x) := \delta(x-c)$. The mean and variance of this "constant random variable"(!) comes out as c and 0, but does this approach have enough mathematical rigor?

Assume that $c > 0$ otherwise $\min(X,c) = c$ which is rather trivial. Then $Y := \min(X,c)$ is a nonnegative variable and we have

\begin{align*} E[Y] &= \int_0^\infty P( Y > t) dt \\ &= \int_0^\infty P( X > t, c > t) dt \\ &= \int_0^\infty P( X > t)1\{ c > t\} dt \\ &= \int_0^c P(X > t)dt \end{align*}

• I didn't know about the use of CDF for calculating the expectation of a non-negative r.v. For the convenience of those like me out there, I just add here the link to a related answer: math.stackexchange.com/questions/64186/… – Libra Jul 21 '15 at 15:24

Discrete random variables have densities that are sums of Dirac delta functions so there is nothing wrong with that.

In general, the problem you are asking is equivalent really to one in options pricing where the value of maximums and mininums are computed all the time.

$$\min(X,c) = X - \max(X-c,0)$$ The second term is the pay-off of a call option. There are formulas for this when $X$ and lognormal and in various other cases. You find them by integrating the density of $X$ explicitly.

• Thanks for the response. The random variable X, in this problem, is continuous and by my approach, I was hoping to translate the whole problem to the continuous domain. – Vectorizer Apr 30 '15 at 22:01
• essentially just integrate the density times $(x-c)$ from $c$ to infinity to get the final term in my expression. – Mark Joshi Apr 30 '15 at 22:02

There is no general formula: you have to use the distribution of $X$. You can do this conveniently by exploiting conditioning:

$$P(Y \leq y)=P(Y \leq y|X \leq y)P(X \leq y) + P(Y \leq y|X > y)P(X > y)$$

Now $Y \leq X$, so if $X \leq y$ then $Y \leq y$ by default. So the first term is just $P(X \leq y)$. For the second term, we break into two cases: $y \geq c$ and $y<c$. For $y \geq c$, if $X>y$ then $Y=c$, so $Y \leq y$. Hence again $P(Y \leq y|X > y)=1$ and we get $P(Y \leq y)=1$ whenever $y \geq c$.

Now you try to deal with the second term in the case $y<c$.

• Thanks but I was concentrating first on the validity of using $\delta$() as a pdf for a constant – Vectorizer Apr 30 '15 at 22:09
• @Vectorizer It's valid but unnecessary: if $X$ is continuous, then when you calculate the CDF this way, you will get a differentiable function (except perhaps at $c$). Then you differentiate it to get the pdf. I would recommend sticking to this approach to avoid various pitfalls that you can encounter when using distribution theory informally. – Ian Apr 30 '15 at 22:13

Preliminaries:

1. $E[X] = \int_0^\infty (1-F_{X}(x))dx$, when X $\geq 0$.
2. Z := min(X, Y) $\Rightarrow F_{Z}(z) = F_{X}(z)+F_{Y}(z) - F_{X,Y}(z,z)$
3. A constant value is modeled as a Random Variable whose pdf is $\delta_{D}(z-c)$ and cdf is $H_{\theta}(z-c)$ where $H_{\theta}()$ is the Heaviside Theta function.

Applying the above, to the question Z := min(X, c), X $\geq 0$ AND X${\perp}$c.

$E[Z] = \int_0^\infty (1-F_{Z}(u))du$ = $\int_0^\infty (1-F_{X}(u)-F_{c}(u)+F_{X}(u)F_{c}(u))du = \int_0^\infty (1-F_{X}(u)-H_{\theta}(u-c)+F_{X}(u)H_{\theta}(u-c))du =\\ \int_0^c(1-H_{\theta}(u-c))(1-F_{X}(u))du + \int_{c^{+}}^\infty(1-H_{\theta}(u-c))(1-F_{X}(u))du =\\ \int_0^c(1-F_{X}(u))du$