Probability density functions (normal distribution)
I was trying out some probability questions when I came upon this one that I'm perplexed by;
Let $X \sim N(0,1)$ and $Y = e^X.$ Find the density function of $Y$.
I know the density function for $N(\mu, \sigma 2)$ is $$f(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\left(−\frac{(x−\mu)^2}{2\sigma^2}\right).$$
Is this the function that i use? and how do I find the density function in terms of $Y?$
$\endgroup$ 11 Answer
$\begingroup$The pdf of a random variable can be found by using its cdf. Let consider \begin{equation} \textbf{P}(Y \le x) = \textbf{P}(\exp(X) \le x) = \textbf{P}(X \le \log x) \end{equation} Since $$ \textbf{P}(X \le \log x) = \int_{-\infty}^{\log x} \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(t-\mu)^2}{2\sigma^2}}dt $$ by differentiating with $x$, we obtain $$ f_Y(x) = \frac{d\mathbf{P}(Y\le x)}{x} = \frac{1}{\sqrt{2\pi}\sigma x}\exp\left(-\frac{(\log x-\mu)^2}{2\sigma^2}\right)$$
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