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Show that $$f(y)=\begin{cases} 2y+\frac32&0<y<\frac12\\ 0&\text{otherwise} \end{cases}$$ is a probability density function and determine the expected value of the corresponding random variable $Y$.

What are the steps to complete this exercise?

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closed as off-topic by 5xum, TheGeekGreek, Parcly Taxel, drhab, Dave Jul 18 '17 at 12:54

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  • $\begingroup$ I can't read your characters! $\endgroup$ – Parcly Taxel Jul 18 '17 at 11:12
  • $\begingroup$ @ParclyTaxel: Sneaky characters... maybe they are from GoT? $\endgroup$ – mathreadler Jul 18 '17 at 11:15
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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Jul 18 '17 at 11:15
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. $\endgroup$ – 5xum Jul 18 '17 at 11:15
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You have to show: $f \ge 0$ on $\mathbb R$, $f$ is integrable and $\int_{ - \infty}^{\infty}f(y)dy=1$.

The expected value is given by $E(Y)=\int_{ - \infty}^{\infty}yf(y)dy$.

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Step 1: Identify the two properties of a probability density function for an absolutely continuous random variable.

$$\forall y\in\Bbb R~(f(y)\geqslant 0)\\ \int_\Bbb R f(y)\operatorname d y = 1$$

Step 2: Show that this function has them.

Step 3: Apply the definition of expectation for such a random variable.

$$\Bbb E(Y) = \int_\Bbb R y~f(y)\operatorname d y$$

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