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Expected value and variance of max{Y_1,Y_2}

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations


    3. 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
    $$V(X)=E(X^2)-(E(X))^2=\frac{11}{6}-\left(\frac{4}{3}\right)^2=\frac{1}{18}$$
     
  2. jcsd
  3. Oct 31, 2016 #2

    mfb

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    2016 Award

    Staff: 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 $.
     
  4. Oct 31, 2016 #3

    Ray Vickson

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    Homework Helper

    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##.
     
    Last edited: Oct 31, 2016
  5. Oct 31, 2016 #4
    The cdf is (x-1)^2?
     
  6. Oct 31, 2016 #5

    Ray Vickson

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    Not for all x, no, but maybe for some x. Anyway, you tell me!
     
  7. Oct 31, 2016 #6
    For x from 1 to 2, (x-1)^2
    For x below 1, it is 0
    For x greater than 2, it is 1
     
  8. Oct 31, 2016 #7

    Ray Vickson

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    Correct. You could write it in LaTeX using the "cases" command, exactly as you did in post #1, like this:
    $$F_X(x) = \begin{cases} 0 &\text{if $x < 1$}\\
    (x-1)^2 & \text{if $1 \leq x < 2$}\\
    1 & \text{if $x \geq 2$}
    \end{cases}
    $$
     
    Last edited: Oct 31, 2016
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