Question

If \(f\) is the density function of a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\), show that the tilted density \(f_{t}\) is the density of a normal random variable with mean \(\mu+\sigma^{2} t\) and variance \(\sigma^{2}\).

Short answer

The tilted density function \(f_t(x)\) is given by: \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2(\mu+\sigma^2 t)x + (\mu+\sigma^2 t)^2 - \sigma^4 t^2}{2\sigma^2}\right)\] After comparing this expression to a standard normal probability density function, we can conclude that the tilted density function of a normal random variable is also a normal random variable with mean \(\mu+\sigma^2 t\) and variance \(\sigma^2\).

Step-by-step solution

Define the normal density function f

The probability density function of a normal random variable with mean \(\mu\) and variance \(\sigma^2\) is given by: \[f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\]

Define the tilted density function f_t

Now we find the tilted density \(f_t(x)\): \[f_t(x) = \frac{1}{Z} f(x) \exp(tx)\] Substitute the expression for \(f(x)\): \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2} + tx\right)\]

Simplify the exponent in the function f_t

To simplify the exponent in \(f_t(x)\), we need to group the terms involving \(x\): \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2\mu x + \mu^2}{2\sigma^2} + tx\right)\] Now, we can write this as: \[f_t(x) = \frac{1}{Z} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2 - 2(\mu+\sigma^2 t)x + (\mu+\sigma^2 t)^2 - \sigma^4 t^2}{2\sigma^2}\right)\]

Compare f_t to a standard normal probability density function

We can now compare the expression for \(f_t(x)\) to the standard normal probability density function: \[g(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x - (\mu + \sigma^2 t))^2}{2\sigma^2}\right)\] We can see that the exponent in \(f_t(x)\) and \(g(x)\) is the same, except for the term \(-\sigma^4 t^2\) that doesn't involve \(x\). This extraneous constant term will be absorbed into the normalization constant \(Z\). Now that we have simplified the expression for \(f_t(x)\), we can conclude that it indeed has a mean \(\mu+\sigma^2 t\) and variance \(\sigma^2\), showing that the tilted density function of a normal random variable is also a normal random variable with the given mean and variance.