Find $\mathbb{E}(X)$ and Var$(X)$ from the c.d.f.

Question

Suppose a child plays outside in the yard. On their own, they come back inside at a random time uniformly distributed on the interval [0,1] (Take the units to be hours.) However, if the child is not back in 50 minutes, their mother brings them in. Let X be the time when they come back in.

1. What is the cumulative distribution function F of X?

2. Find E[X].

3. Find Var[X].

My Attempt

1. The c.d.f. is pretty straight forward from the problem. \begin{equation} F(s)= \begin{cases} 0, & s<0\\ s, & 0\leq s<5/6\\ 1, & s\geq 5/6 \end{cases} \end{equation}

2. Differentiating the c.d.f. gives us the p.d.f. $f(s)=1$ for $0\leq s <5/6$. Thus, our expectation is $$\mathbb{E}(X)=\int^{5/6}_0xdx=0.3472$$

3. To find the variance, we also need $\mathbb{E}(X^2)$ $$\mathbb{E}[X^2]=\int^{5/6}_0x^2dx=0.1929$$ So the variance is $$\text{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2=0.1929-0.3472^2=0.0723$$

Can I get verification on my answers? I am a little skeptical on the expectation because I was told that the expectation for a uniform distribution, Unif[a, b], can be calculated as $\mathbb{E}(X)=\frac{a+b}{2}=\frac{5/6+0}{2}=0.4167$, which is different from what I got using the integral method.

Answer 1

your distribution is not absolutely continuous thus your expectation is wrong. Your random variable has a positive probability mass in $x=5/6$

An easy way to calculate its expectation is the following purple area $=35/72$

...you calculated only the area of the triangle...$=25/72$

Try yourself to reason about my hint and calculate the variance...

Answer 2

Another way to solve this is by ysing the heaviside step function $H(x)$ and the dirac delta function $\delta(x)$:

\begin{align*} F(s)= \begin{cases} 0, & s<0\\ s, & 0\leq s<5/6\\ \frac{5}{6} + \frac{1}{6}\cdot H(x-\frac{5}{6}), & s\geq 5/6 \end{cases} \end{align*}

We can then write the pdf in terms of the dirac delta function, which is\begin{align*}f(s)= \begin{cases} 0, & s<0\\ 1, & 0< s<5/6\\ \frac{1}{6}\delta(x-\frac{5}{6}), & s\geq 5/6\\ \end{cases}\end{align*}

The expectation is better computed as an integral over \begin{align*} \mathbb E [X] &= \int\limits_{x=-\infty}^{0} x\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x\cdot \mathbb P[X=x]\\ &= \int\limits_{x=-\infty}^{0} x\cdot f[X=x]dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot f[X=x]dx+ [X=\frac{5}{6}]\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\ &= \int\limits_{x=-\infty}^{0} x\cdot 0dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot 1 dx+ \frac{5}{6}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\ &= 0 + \frac{25}{36\cdot 2}+ \frac{5}{6}\cdot \frac{1}{6}\cdot 1\\ &= \frac{35}{72} \end{align*}

Similarly, one can solve $\mathbb E [X^2]$

\begin{align*} \mathbb E [X^2] &= \int\limits_{x=-\infty}^{0} x^2\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x^2\cdot \mathbb P[X=x]\\ &= 0 + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot f[X=x]dx+ [X=\frac{5}{6}]^2\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\ &= \int\limits_{x=0}^{\frac{5}{6}} x^2dx+ \frac{5^2}{6^2}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\ &= 0 + \frac{125}{216\cdot 3}+ \frac{25}{36}\cdot \frac{1}{6}\cdot 1\\ &= \frac{200}{648}=\frac{25}{81} \end{align*}