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Suppose we generate a random variable $X$ in the following way. First we flip a fair coin. If the coin is heads, take $X$ to have a $Unif(0,1)$ distribution. If the coin is tails, take $X$ to have a $Unif(3,4)$ distribution.

Find the mean and standard deviation of $X$.

This is my solution. I wanted to check if it's correct or if there's a better approach.

Let $Y$ denote the random variable that is $1$ if the coin lands on a head and $0$ otherwise Firstly $\mathbb{E}(\mathbb{E}(X|Y)) = \mathbb{E}(X)$

Thus $\mathbb{E}(\mathbb{E}(X|Y)) = \frac{1}{2} \cdot \mathbb{E}(X|Y=0) + \frac{1}{2} \cdot \mathbb{E}(X|Y=1) = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{7}{2}=2$

Secondly $\mathbb{V}(X) = \mathbb{E}(\mathbb{V}(X|Y))+\mathbb{V}(\mathbb{E}(X|Y))$

Now $\mathbb{V}(X|Y = 0) = \mathbb{V}(X|Y=1) = \frac{1}{12}$. Thus $\mathbb{E}(\mathbb{V}(X|Y)) = \frac{1}{12}$. Next calculating $\mathbb{V}(\mathbb{E}(X|Y)) = \mathbb{E}(\mathbb{E}(X^2|Y)) - (\mathbb{E}(\mathbb{E}(X|Y)))^2 = (\frac{1}{2} \cdot \frac{1}{4} + \frac{49}{4} \cdot \frac{1}{2}) - (2)^2 = \frac{50}{8} - 4.$

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Your calculation is correct, and is a good way I think. One other approach might be just using the PDF of $X$, using uniform PDF, $\Pi(x)$: $$f_X(x)=\frac{1}{2}\Pi(x)+\frac{1}{2}\Pi(x-3)$$ Expected value can be fairly easy via both method, we just need $E[X^2]$: $$E[X^2]=\frac{1}{2}\int_0^{1}x^2dx+\frac{1}{2}\int_3^4x^2dx=\frac{4^3-3^3+1^3}{6}=\frac{19}{3}$$ which yields $\operatorname{var}(X)=19/3-4=7/3$, as yours.

Note: Add 1/12 to your final answer, since your answer is for $V(E[X|Y])$.

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  • $\begingroup$ wait we get different answers for variance $ 7/3 \neq 50/8-4$ where did i go wrong? $\endgroup$ – Iltl Apr 20 at 14:26
  • $\begingroup$ @Iltl your answer is for V(E[X|Y]), if you add E[V(X|Y)] to that, they become equal. $\endgroup$ – gunes Apr 20 at 15:24
  • $\begingroup$ Gotcha i forgot i didn't include it thanks! $\endgroup$ – Iltl Apr 20 at 15:25
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There are generally two ways to approach these types of problems: by (1) Finding the second stage expectation $E(X)$ with the theorem of total expectation; or by (2) Finding the second stage expectation $E(X)$, using $f_{X}(x)$. These are equivalent methods, but you might find one easier to comprehend, so I present them both in detail below for $E(X)$. The approach is similar for $Var(X)$, so I exclude its presentation, but can update my answer if you really need it.

Method (1) Finding the second stage expectation $E(X)$ with the theorem of total expectation

In this case, the Theorem of Total Expectation states that:

\begin{eqnarray*} E(X) & = & \sum_{y=0}^{1}E(X|Y=y)P(Y=y)\\ & = & \sum_{y=0}^{1}E(X|Y=y)f_{Y}(y) \end{eqnarray*}

So, we simply need to find the corresponding terms in the line above for $y=0$ and $y=1$. We are given the following:

\begin{eqnarray*} f_{Y}(y) & = & \begin{cases} \frac{1}{2} & \text{for}\,y=0\,(heads),\,1\,(tails)\\ 0 & \text{otherwise} \end{cases} \end{eqnarray*}

and

\begin{eqnarray*} f_{X|Y}(x|y) & = & \begin{cases} 1 & \text{for}\,3<x<4;\,y=0\\ 1 & \text{for}\,0<x<1;\,y=1 \end{cases} \end{eqnarray*}

Now, we simply need to obtain $E(X|Y=y)$ for each realization of $y$:

\begin{eqnarray*} E(X|Y=y) & = & \int_{-\infty}^{\infty}xf_{X|Y}(x|y)dx\\ & = & \begin{cases} \int_{3}^{4}x(1)dx & \text{for}\,y=0\\ \int_{0}^{1}x(1)dx & \text{for}\,y=1 \end{cases}\\ & = & \begin{cases} \left.\frac{x^{2}}{2}\right|_{x=3}^{x=4} & \text{for}\,y=0\\ \left.\frac{x^{2}}{2}\right|_{x=0}^{x=1} & \text{for}\,y=1 \end{cases}\\ & = & \begin{cases} \frac{7}{2} & \text{for}\,y=0\\ \frac{1}{2} & \text{for}\,y=1 \end{cases} \end{eqnarray*}

So, substituting each term into the Theorem of Total Expectation above yields:

\begin{eqnarray*} E(X) & = & \sum_{y=0}^{1}E(X|Y=y)f_{Y}(y)\\ & = & E(X|Y=0)f_{Y}(0)+E(X|Y=1)f_{Y}(1)\\ & = & \left(\frac{7}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\\ & = & 2 \end{eqnarray*}

Method (2) Finding the second stage expectation $E(X)$, using $f_{X}(x)$

To use this method, we first find the $f_{X,Y}(x,y)$ and $f_{X}(X)$. To begin, recall that $f_{X,Y}(x,y)$ is given by:

\begin{eqnarray*} f_{X,Y}(x,y) & = & f_{X|Y}(x|y)f_{Y}(y)\\ & = & \begin{cases} \left(1\right)\left(\frac{1}{2}\right) & \text{for}\,3<x<4;\,y=0\\ \left(1\right)\left(\frac{1}{2}\right) & \text{for}\,0<x<1;\,y=1 \end{cases}\\ \end{eqnarray*}

and we can find $f_{X}(x)$ by summing out the $y$ component:

\begin{eqnarray*} f_{X}(x) & = & \sum_{y=0}^{1}f_{X,Y}(x,y)\\ & = & f_{X,Y}(x,0)+f_{X,Y}(x,1)\\ & = & \frac{1}{2}I(3\le x\le4)+\frac{1}{2}I(0\le x\le1) \end{eqnarray*}

And now, we can just find $E(X)$ using the probability density function of $f_{X}(x)$ as usual:

\begin{eqnarray*} E(X) & = & \int_{-\infty}^{\infty}xf_{X}(x)dx\\ & = & \int_{-\infty}^{\infty}x\left[\frac{1}{2}I(3\le x\le4)+\frac{1}{2}I(0\le x\le1)\right]dx\\ & = & \frac{1}{2}\int_{-\infty}^{\infty}xI(3\le x\le4)dx+\frac{1}{2}\int_{-\infty}^{\infty}xI(0\le x\le1)dx\\ & = & \frac{1}{2}\int_{3}^{4}xdx+\frac{1}{2}\int_{0}^{1}xdx\\ & = & \left(\frac{1}{2}\right)\left.\left(\frac{x^{2}}{2}\right)\right|_{x=3}^{x=4}+\left(\frac{1}{2}\right)\left.\left(\frac{x^{2}}{2}\right)\right|_{x=0}^{x=1}\\ & = & \left(\frac{1}{2}\right)\left(\frac{7}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\\ & = & 2 \end{eqnarray*}

the same two approaches can be used to compute $Var(X)$.

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This problem can be simplified substantially by decomposing the random variable of interest as a sum of two independent parts:

$$X = U+3V \quad \quad \quad \quad U \sim \text{U}(0,1) \quad \quad \quad \quad V \sim \text{Bern}(\tfrac{1}{2}).$$

Using this decomposition we have mean:

$$\begin{equation} \begin{aligned} \mathbb{E}(X) = \mathbb{E}(U+3V) &= \mathbb{E}(U) + 3 \mathbb{E}(V) \\[6pt] &= \frac{1}{2} + 3 \cdot \frac{1}{2} = 2, \\[6pt] \end{aligned} \end{equation}$$

and variance:

$$\begin{equation} \begin{aligned} \mathbb{V}(X) = \mathbb{V}(U+3V) &= \mathbb{V}(U) + 3^2 \mathbb{V}(V) \\[6pt] &= \frac{1}{12} + 9 \cdot \frac{1}{4} \\[6pt] &= \frac{1}{12} + \frac{27}{12} \\[6pt] &= \frac{28}{12} = \frac{7}{3}, \\[6pt] \end{aligned} \end{equation}$$

which gives the corresponding standard deviation:

$$\begin{equation} \begin{aligned} \mathbb{S}(X) = \sqrt{\mathbb{V}(X)} &= \sqrt{\frac{7}{3}} \approx 1.527525. \\[6pt] \end{aligned} \end{equation}$$

As you can see, this simplifies the calculations substantially, and does not require the use of iterated expectations or variance.

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Comment: Here is a brief simulation, comparing approximate simulated results with theoretical results derived in this Q and A. Everything below matches within the margin of simulation error.

Also see Wikipedia on Mixture Distributions, under Moments, for some relevant formulas.

set.seed(420)  # for reproducibility
u1 = runif(10^6);  u2 = runif(10^6, 3, 4)
ht = rbinom(10^6, 1, .5)
x = ht*u1 + (1-ht)*u2
mean(x);  2
[1] 2.001059   # aprx E(X) = 2
[1] 2          # proposed exact
var(x); 7/3
[1] 2.332478   # aprx Var(X)
[1] 2.333333
mean(x^2); 19/3
[1] 6.336712   # aprx E(X^2)
[1] 6.333333 

hist(x, br=40, prob=T, col="skyblue2")

enter image description here

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