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A random variable $X$ has the density function

$f(x)= \begin{Bmatrix} c(x+\sqrt{x}) & x\epsilon [0,1]\\ 0& otherwise \end{Bmatrix}.$

(c)  Determine the probability density function of $Y$ $=$ $X^2$
in Probability by Boss | 75 views

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The relationship between the cumulative distribution $F$ and the probability density $f$ is expressed by

$\Large F(a) = P\{ X \in (\infty,a] \} = \int_{- \infty}^{a} f(x).dx$

Density has the same role as the probability mass function for discrete random variables, it tells which values x are relatively more probable for X than others.

Differentiating both sides,

$\Large \frac{d}{da}F(a) = f(a)$

Now coming to the above given question,

$F_Y(a) = P(Y \leq a) = P(X^2 \leq a) \ $$ = P(X \leq \sqrt a) = F_X(\sqrt a)$

$F_Y(a) = F_X(\sqrt a)$

Differentiation gives,

$\large \frac{d}{da} F_Y( a) = \frac{d}{da} F_X( \sqrt a) $

$ f_Y( a)  = $$\Large \frac{f_X(\sqrt a)}{2\sqrt a}$
by Boss
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What is the physical significance of this question?

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