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We have a given mixed CDF:

$$P(X \leq x) = 1- \alpha e^{-\mu(x-1)} \quad \text{if x $\geq 1$}$$

$$P(X \leq x) = 0 \quad \text{if x $< 1$}$$

As you can see, if $P(X=1) = 1- \alpha$

I want to get $E[X]$.

If I differentiate the CDF I get the following PDF:

$$f_x = \alpha \mu e^{-\mu (x-1)} \quad \text{if $x \geq $ 1}$$ $$f_x = 0 \quad \text{if $x < 1$}$$

So should we use the formula: $$E[X] = \int_1^{\infty} x\cdot \alpha \mu e^{-\mu(x-1)}dx$$

To get the expectation, or should we use

$f_x = \alpha \mu \quad \text{x = 1}$

and $$E[X] = \mu\alpha\cdot 1+\int_1^{\infty} x\cdot \alpha \mu e^{-\mu(x-1)}dx$$

kimchi lover
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VLC
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    I think you are trying to say this is a mixture distribution with probability $\alpha$ taken from an exponential distribution above $x=1$ with rate parameter $\mu$ and with probability $1-\alpha$ a point at $x=1$. I would have thought that would make the expectation $\alpha(1+\frac1\mu)+(1-\alpha)1 =1 +\frac{\alpha}{\mu}$ but I have not checked – Henry Mar 09 '23 at 14:39
  • @Henry So the term $\alpha (1 + \frac{1}{\mu})$ is the solution of the integral from 1 to infinity for the expectation. How did you get (1 - \alpha) ? – VLC Mar 09 '23 at 14:58
  • @Henry I got it now. Good answer – VLC Mar 09 '23 at 15:03
  • There is a general expression for the mean in terms of the cdf only: https://math.stackexchange.com/q/2136984/321264. – StubbornAtom Mar 09 '23 at 16:01

2 Answers2

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We can use conditioning:

$$\operatorname{E}[X] = \operatorname{E}[X \mid X = 1]\Pr[X = 1] + \operatorname{E}[X \mid X > 1]\Pr[X > 1]. \tag{1}$$ The conditional distribution of $Y = X \mid X > 1$ has the survival function $$S_Y(y) = \Pr[Y > y] = \Pr[X > y \mid X > 1] = \frac{\Pr[(X > y) \cap (X > 1)]}{\Pr[X > 1]}. \tag{2}$$ If $y > 1$, then the event $(X > y) \cap (X > 1) = (X > y)$. If $y \le 1$, then $(X > y) \cap (X > 1) = (X > 1)$. $$S_Y(y) = \begin{cases} 1, & y \le 1 \\ \frac{S_X(y)}{S_X(1)}, & y > 1 \end{cases} \tag{3}$$ where $$\frac{S_X(y)}{S_X(1)} = \frac{1 - \Pr[X \le y]}{1 - (1-\alpha)} = \frac{\alpha e^{-\mu(x-1)}}{\alpha} = e^{-\mu(x-1)}. \tag{4}$$ Hence $Y$ has density $$f_Y(y) = \begin{cases} 0, & x \le 1 \\ \mu e^{-\mu(x-1)}, & x > 1. \end{cases} \tag{5}$$ It follows that $$\operatorname{E}[X] = (1) (1-\alpha) + \left(\int_{x=1}^\infty x \mu e^{-\mu(x-1)} \, dx \right)(\alpha) = (1-\alpha) + \left(1 + \frac{1}{\mu}\right) \alpha = 1 + \frac{\alpha}{\mu}. \tag{6}$$

That said, we can recognize that this mixed distribution is a location-transformed, zero-inflated exponential distribution with rate $\mu$. The zero inflation factor is $1-\alpha$ and the location parameter is $1$. That is to say, $$X = 1 + BW, \tag{7}$$ where $B \sim \operatorname{Bernoulli}(\alpha)$ and $W \sim \operatorname{Exponential}(\mu)$ are independent. Then $$\operatorname{E}[X] = \operatorname{E}[1 + BW] = 1 + \operatorname{E}[B]\operatorname{E}[W] = 1 + \alpha \cdot \frac{1}{\mu}. \tag{8}$$

heropup
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2

Since the r.v is almost surely positive (in fact $X\geq 1$ a.s) one can use the formula $$ EX=\int_{0}^\infty P(X>x) \, dx. $$

Let $B = \{1\}$.

If you want to use the density, then note that $X$ is absolutely continuous with respect to the measure $\nu = \lambda+\mu$ where $\lambda$ is the Lebesgue measure and $\mu$ is the counting measure via the density $$ g(x) = f(x)I(x>1) + (1-\alpha)I(x=1) $$ where $f(x) = \alpha \mu e^{-\mu (x-1)} $since $$ \begin{align} \int_{A} f(x)d \nu &=\int_{A}f(x)\, d\lambda+\int_{A} f(x) d \mu\\ &=\int_{A\cap B^c} f(x)\,d\lambda+ \int_{A\cap B} f(x)\,d\lambda + \int_{A\cap B^c} f(x)\,d\mu+ \int_{A\cap B} f(x)\,d\mu \\ &=\int_{A\cap B^c} f(x)\, d\lambda+(1-\alpha)\delta(1\in A)\\ &=P(X\in (1, \infty)\cap A)+P(X\in \{1\}\cap A)\\ &=P(X\in A) \end{align} $$ for any Borel set $A$.

So $$ EX =\int xg(x)d\nu(x) =\int_{1}^\infty xf(x)\, dx + 1-\alpha =\alpha\int_{1}^\infty \mu e^{-\mu (x-1)}+(1-\alpha)=1+\frac{\alpha}{\mu} $$

Sri-Amirthan Theivendran
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