Question

A filtration process removes a random proportion of particulates in water to which it is applied. Suppose that a sample of water is subjected to this process twice. Let x1 be the proportion of the particulates that are removed by the first pass. Let X2 be the proportion of what remains after the first pass that is removed by the second pass. Assume that X1 and X2 are independent random variables with common pdf. f(x) = 4x3, for 0 < x <1 and f(x) = 0 otherwise. Let Y be the proportion of the original particulates that remain in the sample after two passes. Then Y = (1 - X1)(1 - X2). Find E(Y).

Answer 1

Answer:

$E(Y)=\frac{1}{25}$

Step-by-step explanation:

Let's start defining the random variables for this exercise :

$X_{1}:$ '' The proportion of the particulates that are removed by the first pass ''

$X_{2}:$ '' The proportion of what remains after the first pass that is removed by the second pass ''

$Y:$ '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

$Y=(1-X_{1})(1-X_{2})$

We also assume that $X_{1}$ and $X_{2}$ are independent random variables with common pdf.

The probability density function for both variables is $f(x)=4x^{3}$ for $0$ and $f(x)=0$ otherwise.

The first step to solve this exercise is to find the expected value for $X_{1}$ and $X_{2}$.

Because the variables have the same pdf we write :

$E(X_{1})= E(X_{2})=E(X)$

Using the pdf to calculate the expected value we write :

$E(X)=\int\limits^a_b {xf(x)} \, dx$

Where $a=$ ∞ and $b=$ - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between $0$ and $1$ ⇒

Using the pdf we calculate the expected value :

$E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}$

⇒ $E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}$

Now we need to use some expected value properties in the expression of $Y$ ⇒

$Y=(1-X_{1})(1-X_{2})$ ⇒

$Y=1-X_{2}-X_{1}+X_{1}X_{2}$

Applying the expected value properties (linearity and expected value of a constant) ⇒

$E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})$

Using that $X_{1}$ and $X_{2}$ have the same expected value $E(X)$ and given that $X_{1}$ and $X_{2}$ are independent random variables we can write $E(X_{1}X_{2})=E(X_{1})E(X_{2})$   ⇒

$E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})$ ⇒

$E(Y)=E(1)-2E(X)+[E(X)]^{2}$

Using the value of $E(X)$ calculated :

$E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}$

$E(Y)=\frac{1}{25}$

We find that the expected value of the variable $Y$ is $E(Y)=\frac{1}{25}$