# Is a variable transformed from a random variable a random variable?

Let a random variable $X$ be uniform distribution in $(0,1)$ and another variable $Y=X^2$. I calculate the sample space of $Y$ is also $(0,1)$, and the cdf and pdf of $Y$ are $F_Y(y)=\sqrt{y}$ and $f_Y(y) = \frac{1}{2\sqrt{y}}$. I think $Y$ should also be a random variable and the integration of pdf of $Y$ should give 1 but it does not seems right. I don't expect the following integration gives me 1. $$\int^1_0{\frac{1}{2\sqrt{y}}}dy = -\frac1 4 y^{-\frac 3 2}|^1_0$$

Did I calculate the sample space and pdf and the integration correctly? Or a variable derived from a random variable is not necessarily a random variable.

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You incorrectly computed the anti-derivative. Remember that $\int f_Y(y) \mathrm{d}y = F_Y(y) + c$. – Sasha Aug 31 '12 at 16:08
Yes a function of a random variable is a random variable and did and Andre have shown you the correct answer. – Michael Chernick Aug 31 '12 at 16:24

## 2 Answers

At the end, there was a slip in the integration. The definite integral is $1$.

Note that an antiderivative of $\frac{1}{2\sqrt{y}}$ is $\sqrt{y}$. (Instead of integrating , that is, reversing the step that got you from $\sqrt{y}$ to $\frac{1}{2\sqrt{y}}$, you differentiated.)

To integrate, use the fact that $\frac{1}{2\sqrt{y}}=\frac{1}{2}y^{-1/2}$. So by the usual integration formula for powers, the integral of this is $\frac{1}{2}\frac{y^{-1/2+1}}{-1/2+1}+C$.

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$$\int^1_0{\frac{1}{2\sqrt{y}}}dy = +\sqrt{y}\,{\Large|}^1_0=1$$

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just a bit too late :-) (+1) – robjohn Aug 31 '12 at 16:10