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I am trying to find the probability distribution of a continuous random variable X which is defined by 𝑃(X=x)=𝑘(2−x)(x+1),where 0≤=x≤2

Can anybody help me to find x and mean expected value E(x) ?

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    $\begingroup$ First of all you should evaluate the value of $k$. Use that $\int\limits_{0}^2 f(x) \, dx =1$ $\endgroup$ Commented Dec 26, 2020 at 15:36
  • $\begingroup$ I suppose that $P(X=x)$ in the above really should be $f_X(x)$ (the probability density function of $X$), as for a continuous random variable we have that $P(X=x)=0$. $\endgroup$ Commented Dec 26, 2020 at 15:45

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You can't find $x$, the question doesn't make sense and I suspect you meant find $k$ so I will proceed with that A continuous random variable must satisfy the property $\int f(x) dx=1$ or in other words: the sum of all probability must equal to $1$.

To find $k$ we need $k\times\int_0^2(2-x)(x+1)dx=1$, we have $\int_0^2(2-x)(x+1)dx=\frac{10}{3}$ (this was done quickly on WolframAlpha as the goal here isn't to teach you integration) so we have $k\times \frac{10}{3}=1$ which gives us $k=\frac{3}{10}$

For your second question, we have $E(X)=\int xf(x)$ so to find the expected value you should evaluate $\frac{3}{10}\int_0^2x(2-x)(x+1) dx$ which gives $\frac{4}{5}$

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    $\begingroup$ @projectilemotion yes, thank you for pointing it out, op asked as similar question about a discrete random variable like an hour ago so I just copied the initial sentence and forgot this part $\endgroup$
    – Sergio
    Commented Dec 26, 2020 at 15:48
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Your question holds a couple of mistakes. First, if you are dealing with continuous random variable then, for any x the probability P(X=x) is equal to 0 by definition. Secondly, the task to find $x$ makes no sense, since $x$ is a name of an argument here.

I would assume that by P(X=x) you mean a probability density function (PDF) $f_X(x)$, which is also defined as 0 everywhere outside $x\in [0,2]$, and will build my answer based on that information.

  1. Find $k$. For a PDF following conditions hold by definition
  • $f_X(x)\geq0$ for all $x$ (clearly holds);
  • $\int_{-\infty}^{\infty} f_X(x)dx=1$

Calculating the integral above gives us the equation $\frac{10}{3}\cdot k=1$, thus $k=\frac{3}{10}$.

  1. calculate $E[X]$. By definition, $$E[X]=\int_{-\infty}^{\infty} xf_X(x)dx = \int_{0}^{2} x\cdot \frac{3}{10}(2-x)(x+1)dx=\frac45$$
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