Question

"Consider a nonlinear amplifier whose input X and the output Y are related by its transfer characteristic: ​​​​​​​ Assuming that X ∼ N(0, 1) compute the pdf of Y ." Y={ X^ 1/2 , X≥0 ; −∣X∣^ 1/2 , X<0 ​

Answer 1

The probability density function (pdf) of Y is given by:

`\[ f_Y(y) = (1)/(4)e^{-(y^2)/(2)}, \quad y \in \mathbb{R} \]`

Step-by-step explanation:

The transfer characteristic of the nonlinear amplifier relates the input \(X\) to the output \(Y\) through a piecewise function. When
`\(X \geq 0\), \(Y = √(X)\), and when \(X < 0\), \(Y = -√(|X|)\). To find the pdf of \(Y\), we need to determine the distribution of \(Y\) given the distribution of \(X\).`

First, consider the case when
`\(X \geq 0\). In this case, \(Y = √(X)\). To find the pdf of \(Y\), we use the probability density function of \(X\), which is given as \(f_X(x) = (1)/(√(2\pi))e^{-(x^2)/(2)}\) for \(X \sim N(0, 1)\). Applying the transformation \(Y = √(X)\), we use the formula for the pdf of a transformed random variable to obtain the pdf of \(Y\).`

Next, when
`\(X < 0\), \(Y = -√(|X|)\). Again, we use the pdf of \(X\) and the transformation formula to find the pdf of \(Y\) in this case. Finally, by considering both cases, we arrive at the final pdf of \(Y\) as a combination of the pdfs for \(Y\) when \(X \geq 0\) and \(X < 0\), resulting in the provided expression.`