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I am having trouble with finding P(X=1). I am not sure how you would find this since 1 is not included. Also, I believe my cdf is wrong since they do not add up to 1.

The problem states, A continuous random variable defined on the interval [0,2] has the following mixed density function:

f(x)= \begin{cases} x &\text{if } 0 \leq x<1\\ \frac{-1}3 x + \frac{2}3 &\text{if } 1< x \leq 2 \end{cases}

Use the Law of Total Probability to find P(X=1).

I found the cumulative distribution function by integrating the mixed density function that was provided.

F(x)= \begin{cases} \frac{1}2 &\text{if } 0 \leq x<1\\ \frac{1}6 &\text{if } 1< x \leq 2 \end{cases}

1 = P(X=1) by the law of total probability, but wouldn't P(x=1) = 0 since 1 is not included.

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  • $\begingroup$ To begin with, your PDF f(x) does not integrate to 1. $\endgroup$ – kludg Apr 12 '17 at 3:11
  • $\begingroup$ How would I set up the integration? I thought I have to integrate from 0 to 1 and from 1 to 2. $\endgroup$ – Meg Apr 12 '17 at 3:26
  • $\begingroup$ @Meg, your PDF is integrate to $\frac12+\frac16\neq 1$. So this is not PDF. Moreover, if the distribution as absolutely continuous, then $\mathbb P(X=1)=0$. And finally, CDF $F(x)\neq P(X=x)$. Check your question. $\endgroup$ – NCh Apr 12 '17 at 3:32
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    $\begingroup$ If I'm understanding the meaning of the term "mixed density function" correctly here, it sounds like $X$ is meant to be distributed according to a distribution with partially discrete and partially continuous properties. In this case, there is a concentration of nonzero probability at exactly $X = 1$, with the remaining probability distributed continuously according to the given distribution. Since the given continuous part of the PDF integrates to $2/3$, that leaves a probability of $1/3$ to be concentrated at $X=1$. So $P(X=1) = 1/3$. $\endgroup$ – John Barber Apr 12 '17 at 4:15
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    $\begingroup$ @NCh I'm not sure I follow you. I think the mixed distribution you describe there wouldn't work because the continuum portion of it already includes $0$ since it says $0\leq x < 1$. I believe the distribution of $X$ in the original post is implied to be: \begin{equation} f(x) = \frac{1}{3} \delta(x-1) \;+\; \begin{cases} x & 0 \leq x < 1\\ -\frac{1}{3}x + \frac{2}{3} & 1 < x \leq 2 \end{cases} \end{equation} $\endgroup$ – John Barber Apr 12 '17 at 4:46

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