Question

Let X and Y be independent identically distributed (IID) random variables with the exponential probability density functions fX (w)=f Y (w)=λe⁻λʷ u(w) where u(w) is the unit-step function. (a) Determine the probability density function for the ratio 0≤R= X / X+Y ≤1 i.e. f_R (r),0≤r≤1. (b) Let A be the event X<1/Y. Determine the conditional pdf of X given that A occurs and that Y=y; that is, determine fX (x∣A,Y=y).

Answer 1

To find the pdf of the ratio R = X / (X+Y), we first find the cumulative distribution function (cdf) of R and differentiate it to obtain the pdf. Using the exponential pdfs of X and Y, we find that the cdf of R is (1 - e^(-λr))^2. Differentiating the cdf with respect to r gives the pdf of R as 2λe^(-λr)(1 - e^(-λr)).

Step-by-step explanation:

In order to determine the probability density function (pdf) for the ratio R = X / (X+Y) where X and Y are independent identically distributed (IID) random variables with exponential pdfs, we need to find the cumulative distribution function (cdf) for R and differentiate it to obtain the pdf.

First, we find the cdf of R:
P(R ≤ r) = P(X / (X+Y) ≤ r) = P(X ≤ r(X+Y)) = P(X ≤ rX + rY)
Since X and Y are independent, we can rewrite the cdf as:
P(R ≤ r) = P(X ≤ rX)P(Y ≤ rY) = P(X ≤ r)P(Y ≤ r)

Using the exponential pdfs of X and Y, we have:
P(X ≤ r) = ∫[0,r] λe^(-λx) dx = 1 - e^(-λr)
P(Y ≤ r) = ∫[0,r] λe^(-λy) dy = 1 - e^(-λr)

Therefore, the cdf of R is:
P(R ≤ r) = (1 - e^(-λr))^2
To find the pdf of R, we differentiate the cdf with respect to r:
f_R(r) = d/d_r[(1 - e^(-λr))^2] = 2λe^(-λr)(1 - e^(-λr))