Question

\(X\) is a normal distribution with \(N(0,2)\) and PDF \(f_X\). \(Z\) is standard normal distribution with \(N(0,1)\) and PDF \(f_Z\). If \[f_X(2) = \frac{1}{2}f_Z(x_\circ)\], what is \(x_\circ\)?

  1. 1
  2. 2
  3. None of the above

Answer

Correct answer: \(x_\circ=1\)

Let CDF of standard normal distribution, \(Z\) be \(\Phi(x)\). Any other normal random variable with mean \(\mu\) and standard deviation \(\sigma\) can be written as \(X = \sigma Z + \mu\). So CDF of \(X\) (denoted by \(F_X(x)\)):

\[\begin{aligned} F_X(x) &= P(X \leq x) \\ F_X(x) &= P(\sigma Z + \mu \leq x) \\ F_X(x) &= P(Z \leq \frac {x - \mu}{\sigma}) \\ F_X(x) &= \Phi(\frac {x - \mu}{\sigma}) \end{aligned} \]

and PDF of X can be find by differentiating \(F\). \[\begin{aligned} f(x) &= \frac {d}{d(x)} F_X(x) \\ f(x) &= \frac {d}{d(x)} \Phi(\frac {x - \mu}{\sigma}) \\ f(x) &= \frac {1}{\sigma} \Phi'(\frac {x - \mu}{\sigma}) \\ f(x) &= \frac {1}{\sigma} f_Z(\frac {x - \mu}{\sigma}) \end{aligned} \]

So for the above question, it is given for \(X\) that \(\mu = 0\) and \(\sigma = 2\), plugging in values: \[\begin{aligned} f(x) &= \frac {1}{2}f_Z(\frac {x}{2}) \\ f_X(2) &= \frac {1}{2}f_Z(\frac {2}{2}) = \frac{1}{2}f_Z(x_\circ) \\ &\Rightarrow x_\circ = 1 \end{aligned} \]

In R:

dnorm(2,0,2)
## [1] 0.1209854
0.5*(dnorm(1,0,1))
## [1] 0.1209854

Thanks for reading. If you like the question, how about some love and coffee: Buy me a coffee