Calculate the densitiy of the random variable $X$

Given is $U \sim R(0,1)$ and $X=g(U) = -\frac{1}{\lambda} \ln(U), \space \space \lambda > 0$

What's the density of random variable $X$?

$X \sim R(a,b)$ in general mean equal distribution.

Density for eual distribution is defined: $$P(a \leq X \leq b) = \int_{a}^{b} f(x) dx = \frac{1}{b-a} \int_{a}^{b} dx = 1$$

Then in this example, we have $$P(0 \leq X \leq 1) = \int_{0}^{1} g(U) dU = \int_{0}^{1} -\frac{1}{\lambda} \ln(U) \space\space dU = -\frac{1}{\lambda}\int_{0}^{1} \ln(U) \space dU = -\frac{1}{\lambda} \left[U \cdot \ln(U)-U\right]_{0}^{1}=$$

But now you see there will be problem with $\ln(0)$ when I insert intervals for calculate it. No defined calculation..

What is mistake? I don't see it :(

• What your integral calculates is the expectation of $X$, not the density function of $X$. Your last integral can be re-expressed using the change of variables $U=\exp (-V)$; how does your concern about the $U=0$ case translate into $V$ terms? – kimchi lover Dec 6 '17 at 13:03

Use CDF transformation method. $$P(X\le x) = P(-\frac{1}{\lambda}.\ln U \le x)$$

$$= P(\ln U \ge -\lambda x)$$

$$= P(U \ge e^{-\lambda x})$$

$$F_{X}(x) = \int_{e^{-\lambda x}}^{1} dU = 1-e^{-\lambda x}$$

$$f(x) = \frac{d}{dx} (1-e^{-\lambda x})$$

$$f(x) = \lambda.e^{-\lambda x}$$

Hence X is $exp(\lambda)$.

• Oh I dident know this formula thank you very much for show how do it good. Then no wonder I get this no defined result because I use wrong formula.. I have other question, what when random variable would be $\sqrt{X}$ ? Can I do it like you and in the end take square root of solution? I don't know when this make sense I only ask for interested. – roblind Dec 6 '17 at 13:16
• You could do the same for $Y = \sqrt{X}$. with some f(x) and you want to find f(y). – Satish Ramanathan Dec 6 '17 at 13:18
• and you are welcome – Satish Ramanathan Dec 6 '17 at 13:18
• Ok I think I make next question for this and solve it with your idea. Thank you again :) – roblind Dec 6 '17 at 13:28