Question I was given: Let V be a uniform random variable distributed over the interval (0,1). Let $\ X = \frac{1}{\sqrt(U)}$. What is the cumulative distribution function and probability density function of X?
I understand the basic concept here behind cumulative distribution functions and probability density functions. What's throwing me off here is the additional parameter saying that V is a uniform random variable.
Is the cumulative distance function just: $\ \int\limits_0^1 \frac{1}{\sqrt(V)}dV $ which would then just equal 2?
Then the prob dens function would be: $$ \frac{1}{\sqrt(V)}-----0<V<1 $$ $$ 0------otherwise $$ This seems too simple to me to be the correct answer though. Am I missing something? Am I finding the cumulative distance function of V rather than of X?
We find the cumulative distribution function $F_X(x)$ of $X$. So we want to find $\Pr(X\le x)$.
If $x\le 1$, then $F_X(x)=0$. For since $0\lt U\lt 1$, it is impossible for $\dfrac{1}{\sqrt{U}}$ to be $\le 1$.
Now we find $F_X(x)$ for $x\gt 1$.
We have $X\le x$ if and only if $\dfrac{1}{\sqrt{U}}\le x$. Flip it over. So $X\le x$ if and omly if $\sqrt{U}\ge \dfrac{1}{x}$, that is, if and only if $U\ge \dfrac{1}{x^2}$.
But $\Pr\left(U\ge \dfrac{1}{x^2}\right)=1-\Pr\left(U\le \dfrac{1}{x^2}\right)$. Since $U$ is uniform, this is simply $1-\dfrac{1}{x^2}$. Thus for $X\gt 1$, $$F_X(x)= 1-\frac{1}{x^2}.$$
For the density function $f_X(x)$, differentiate the cdf.