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Let me be clear, this question is straight from my homework sheet.

Consider $$ K(ω) = \begin{cases} U &\text{for }ω ∈ [0,1/3],\\ 0 &\text{for }ω ∈ (1/3,2/3]\\ D &\text{for }ω ∈ (2/3,1]\end{cases} $$ Find U and D for the following expected value and standard deviation, $\mathbb{E}(X) = 0.1$, $\sigma = 0.2$.

  1. So I know how to integrate and find the expected value and the variance. The problem is how am I supposed to integrate for intervals (1/3,2/3] since the value of 1/3 is excluded from the interval, also the interval is not given for K(ω)=D, then how can we even find its value ask asked.
  2. Can someone please provide a step by step solution here, I know the fact that a CDF can take values between 0 and 1, but I am confused on how to apply these facts and integration on this sum.
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  • $\begingroup$ The definition of $K(\omega)$ does not make sense. I guess it should be $K(\omega)=D, \omega\in(2/3,1]$. If that is the case, then all you need to do is to calculate the expectation and variance of $\omega$ in terms of the unknowns $U$ and $D$ and then set them equal to the given values. $\endgroup$ – Moss Murderer Jan 19 '18 at 17:07
  • $\begingroup$ K(ω)=D,ω∈(2/3,1]; how should I integrate this one? I can integrate K(ω) = U from [0,1/3], but what about K(ω) = D ? Could you please try to solve this sum. I need to just look at the order in which things are being computed. Thanks for your comment $\endgroup$ – Kir'Shara Jan 19 '18 at 18:04
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Given PDF, $K(\omega)=\begin{cases} U, & \text{$0\le\omega\le1/3$}.\\ 0, & \text{$1/3<\omega\le2/3$}.\\ D, & \text{$2/3<\omega\le1$}.\\ \end{cases}$

$$ \begin{align} E(\omega)&=\int_0^1\omega K(w)d\omega \\ &=U\int_0^{1/3}\omega d\omega\quad +\quad 0\int_{1/3}^{2/3}\omega d\omega\quad +\quad D\int_{2/3}^{1}\omega d\omega\\ &=\frac{U+5D}{18}=0.1 \end{align} $$

Likewise, you can express $Var(\omega)=\int_0^1(\omega-E(\omega))^2 K(w)d\omega$ in terms of $U$ and $D$. And then solve the system of two equations to get the values of $U$ and $D$. There is no need to compute CDF.

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  • $\begingroup$ Thank You sir. I finally understand the order now. I had been confused between integration of open intervals, but it’s solved now. Thanks. $\endgroup$ – Kir'Shara Jan 19 '18 at 18:52

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