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

Question

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?

Answer 1

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*}

Answer 2

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 $y<c$.

Answer 3

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.

Answer 4

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.

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

Answer 5

(1) When $c < 0$, we have $Y=c$, since $X\geq 0 \implies E[Y]=E[c]=c$

(2) When $c \geq 0$, let's define $A=\{X:X\leq c\}$ and $I_A$ be the indicator function for $A$, then we have $A^c=\{X:X>c\}$ and

$Y=XI_A+cI_{A^c}$

$\implies E[Y]=E[XI_A]+cE[I_{A^c}]$, by linearity of expectation

$\implies E[Y]=\int\limits_{0}^{c} xf_{X}(x)dx +cP(X>c)$, by definition of expectation of indicator and $X \geq 0$,

where $f_X(.)$ is the pdf of the r.v. $X$.

Combining (1) and (2), we have,

$E[Y] = \left\{\begin{array}{lr} c, & \text{for } c< 0\\ \int\limits_{0}^{c} xf_{X}(x)dx +cP(X>c), & \text{for } c \geq 0\\ \end{array}\right\}$