# Expected value and variance of max{Y_1,Y_2}

## Homework Statement

Let Y_1,Y_2 be independent random variable with uniform distribution on the interval [1,2]. Define X=max{Y_1,Y_2}. Find p.d.f., expected value and variance.

## The Attempt at a Solution

Since $X=\max\{Y_1,Y_2\}$, this tells $Y_1$ and $Y_2$ must at most $x$. Thus, $\p(\max\{Y_1,Y_2\}\leq x)=\p(\{Y_1\leq x,Y_2\leq x\})=(2-x)^2$ with $1\leq x\leq 2$. So, the $c.d.f.$ of $X$ is
$$\p(x)= \begin{cases} 0&\text{if x<1}\\ (2-x)^2 & \text{if 1\leq x\leq 2}\\ 1& \text{if x>2} \end{cases}$$ where $a,b\in[1,2]$.
Then, the $p.d.f.$ is
$$p(x)= \begin{cases} 2(2-x)&\;\;\text{if 1\leq a,b\leq 2}\\ 0&\;\;\text{otherwise} \end{cases}$$
The expected value of $X$ is
$$E(X)=\int_{1}^{2}(4x-2x^2)\,dx=\frac{4}{3}$$
We also have
$$E(X^2)=\int_{1}^{2}(4x^2-2x^3)\,dx=\frac{11}{6}$$
Therefore, the variance is
$$V(X)=E(X^2)-(E(X))^2=\frac{11}{6}-\left(\frac{4}{3}\right)^2=\frac{1}{18}$$

mfb
Mentor
There is a sign error in the derivation of the cdf.

This is also visible in the expectation value: The maximum of two variables cannot have an expectation value below the expectation value of its arguments (which is 3/2 > 4/3).

By the way, you can use inline LaTeX with double # instead of single $. Ray Vickson Science Advisor Homework Helper Dearly Missed ## Homework Statement Let Y_1,Y_2 be independent random variable with uniform distribution on the interval [1,2]. Define X=max{Y_1,Y_2}. Find p.d.f., expected value and variance. ## Homework Equations ## The Attempt at a Solution Since$X=\max\{Y_1,Y_2\}$, this tells$Y_1$and$Y_2$must at most$x$. Thus,$\p(\max\{Y_1,Y_2\}\leq x)=\p(\{Y_1\leq x,Y_2\leq x\})=(2-x)^2$with$1\leq x\leq 2$. So, the$c.d.f.$of$X$is $$\p(x)= \begin{cases} 0&\text{if x<1}\\ (2-x)^2 & \text{if 1\leq x\leq 2}\\ 1& \text{if x>2} \end{cases}$$ where$a,b\in[1,2]$. Then, the$p.d.f.$is $$p(x)= \begin{cases} 2(2-x)&\;\;\text{if 1\leq a,b\leq 2}\\ 0&\;\;\text{otherwise} \end{cases}$$ The expected value of$X\$ is
$$E(X)=\int_{1}^{2}(4x-2x^2)\,dx=\frac{4}{3}$$
We also have
$$E(X^2)=\int_{1}^{2}(4x^2-2x^3)\,dx=\frac{11}{6}$$
Therefore, the variance is
$$V(X)=E(X^2)-(E(X))^2=\frac{11}{6}-\left(\frac{4}{3}\right)^2=\frac{1}{18}$$
Your ##p(x)## is incorrect, because ##\{ \max(Y_1,Y_2) \leq x \} = \{ Y_1 \leq x \; \& \; Y_2 \leq x \}##, so ##P(X \leq x ) = P(Y_1 \leq x) \cdot P(Y_2 \leq x)## for independent ##Y_1,Y_2##.

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Your ##p(x)## is incorrect, because ##\{ \max(Y_1,Y_2) \leq x \} = \{ Y_1 \leq x \; \& \; Y_2 \leq x \}##, so ##P(X \leq x ) = P(Y_1 \leq x) \cdot P(Y_2 \leq x)## for independent ##Y_1,Y_2##.
The cdf is (x-1)^2?

Ray Vickson
Homework Helper
Dearly Missed
The cdf is (x-1)^2?

Not for all x, no, but maybe for some x. Anyway, you tell me!

Not for all x, no, but maybe for some x. Anyway, you tell me!
For x from 1 to 2, (x-1)^2
For x below 1, it is 0
For x greater than 2, it is 1

Ray Vickson
$$F_X(x) = \begin{cases} 0 &\text{if x < 1}\\ (x-1)^2 & \text{if 1 \leq x < 2}\\ 1 & \text{if x \geq 2} \end{cases}$$