Prove $\frac{e^{t_xX_1}}{\varphi(t_x)}$ is a probability density function

Question

Let $X_1$ be a random variable. I need to prove that the function $f(x_1) = \frac{e^{t_xX_1}}{\varphi(t_x)}$ is a density function, whereby $\varphi(t)$ is the moment generating function of $X_1$ and $t_x$ is the value of $t$ for which $g(t) = tx - \log \varphi(t), x> 0$ is maximised. In other words, $$x = (\log \varphi)'(t_x) \implies x = \frac{\mathbb{E}[X_1e^{t_xX_1}]}{\mathbb{E}[e^{t_xX_1}]}. $$We assume that all moments of $X_1$ are finite, $\varphi(t) = \mathbb{E}[e^{tX_1}]< \infty, \forall t \in \mathbb{R}$. In the book I am reading one finds the following comment: "If the distribution of $X_1$ is transformed with the density $\frac{e^{t_xX_1}}{\varphi(t_x)},$ then the new distribution will have the expectation $x."$ Can somebody understand what is the exact transformation that is meant and why $$\int_{\mathbb{R}} f(x_1)dx_1 = \int_{\mathbb{R}}\frac{e^{t_xX_1}}{\varphi(t_x)}dx_1 = 1.$$ As far as I can see, one may mean the following: we may write $ x = \mathbb{E}[X_1\frac{e^{t_xX_1}}{\mathbb{E}[e^{t_xX_1}]}] = \mathbb{E}[Y_1],$ whereby we set for the pdf of $Y_1$ to be $f_{Y_1}(y_1) = \frac{e^{t_xY_1}}{\mathbb{E}[e^{t_xY_1}]}.$ In that case,$$\mathbb{E}[Y_1] = \int_{\mathbb{R}} Y_1\frac{e^{t_xY_1}}{\mathbb{E}[e^{t_xY_1}]}dy_1 = x.$$ But I am not sure. In particular, I do not see how $\int_{\mathbb{R}} f(x_1)dx_1 = 1.$

Answer 1

Transforming with density means pointwise multiplication with the original density

Start again with your formula $x=\mathbb E\bigl[X_1\frac{e^{t_xX_1}}{\mathbb E[e^{t_xX_1}]}\bigr] = \mathbb E[Y_1]$, where $Y_1$ should be the transformed random variable. If we write the left expectation as an integral, we see $$ \mathbb E\bigl[X_1\frac{e^{t_xX_1}}{\mathbb E[e^{t_xX_1}]}\bigr] = \int_{\mathbb R} x_1\frac{e^{t_xx_1}}{\mathbb E[e^{t_xx_1}]} \mathbb P_{X_1}(\mathrm dx_1), $$ or, if $X_1$ has a density $f_{X_1}$, $$ \mathbb E\bigl[X_1\frac{e^{t_xX_1}}{\mathbb E[e^{t_xX_1}]}\bigr] = \int_{\mathbb R} x_1\frac{e^{t_xx_1}}{\mathbb E[e^{t_xX_1}]} f_{X_1}(x_1)\mathrm dx_1 = \int_{\mathbb R} x_1 f(x_1) f_{X_1}(x_1)\mathrm dx_1. $$ You see, the density of $Y_1$ is not $f(x_1)$, but rather the product $f(x_1)f_{X_1}(x_1)$. And this is really a pdf: $$ \int_{\mathbb R} f(x_1) f_{X_1}(x_1)\mathrm dx_1 = \mathbb E\left[f(X_1)\right] = \mathbb E\bigl[\frac{e^{t_xX_1}}{\mathbb E[e^{t_xX_1}]}\bigr] = 1. $$ If $X_1$ has no density, it works the same way, but we have to work with the measures themselves, $\mathbb P_{Y_1}(\mathrm dx_1) = f(x_1)\mathbb P_{X_1}(\mathrm dx_1)$. This is a probability measure because $$ \int_{\mathbb R} \mathbb P_{Y_1}(\mathrm dx_1) = \int_{\mathbb R} f(x_1)\mathbb P_{X_1}(\mathrm dx_1) = \mathbb E[f(X_1)] = 1. $$

Answer 2

In the definition of $f_1$ you miss another probability density $f$. Let me change a bit your confusing notations.

$$f_1(x)=\frac{e^{tx}}{\varphi(t)}f(x)dx$$ where $\varphi(t)=\int_{R}e^{tx}f(x)dx.$ If $X$ has density $f$ and $X_1$ has density $f_1$ then the mean $m_1$ of $X_1$ is

$$m_1=\int_{R}x_1f_1(x_1)dx_1=\frac{\varphi'(t)}{\varphi(t)}=\frac{d}{dt}\log \varphi(t).$$