Expected value and variance of max{Y_1,Y_2}

[HaLAA]

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
$$V(X)=E(X^2)-(E(X))^2=\frac{11}{6}-\left(\frac{4}{3}\right)^2=\frac{1}{18}$$

 

[mfb]

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]

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 $Y_1,Y_2$.

 

[HaLAA]

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]

The cdf is (x-1)^2?

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

 

[HaLAA]

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]

For x from 1 to 2, (x-1)^2
For x below 1, it is 0
For x greater than 2, it is 1

Correct. You could write it in LaTeX using the "cases" command, exactly as you did in post #1, like this:
$$F_X(x) = \begin{cases} 0 &\text{if $x < 1$}\\
(x-1)^2 & \text{if $1 \leq x < 2$}\\
1 & \text{if $x \geq 2$}
\end{cases}
$$