Probability NPTEL assignment

Question

Find Value of alpha?

Comments

An easy way would be to look at the definition of uniform probability. If the distribution is defined between $(-\alpha,\alpha)$ then all points outside this range has probability $0$. There are only 2 possibilities to consider. One where $\alpha\le1$ and $\alpha\gt 1$. If $\alpha\le1$ then it wouldn't be possible to get $P(|x|\lt1)=P(|x|>1)$. For $\alpha\gt 1$, this would be only possible if there are 4 equal regions i.e.$ [-2,-1],[1,0],[0,1],[1,2]$. So, $\alpha=2$.

Answer 1

taking assumption that $|\alpha |>1$

$P(|x|<1) = \int_{-1}^{1}\frac{1}{2\alpha }dx = $$\frac{1}{\alpha }$

$P(|x|>1) = \int_{-\alpha }^{-1}\frac{1}{2\alpha }dx+\int_{1}^{\alpha }\frac{1}{2\alpha }dx=$$\frac{-1+\alpha }{2\alpha } +\frac{\alpha -1}{2\alpha }= \frac{\alpha -1}{\alpha }$

$\Rightarrow \frac{\alpha -1}{\alpha }=\frac{1}{\alpha }$

$\Rightarrow \alpha =2$

Comments

how r u applying limits in the integration part???

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if |x|<1, it means x ranges from (-1,1),

similiarly if |x|>1 , it means x ranges from (-alpha, -1) and (1, alpha),

x will not range in (-infinity, -alpha) and (alpha, +infinity), because x is distributed over (-alpha, +alpha) only, outside this interval density function=0