Question

4. [20 points) Continuous Cate and Digital Dave independently pick random numbers C and D, where has the continuous pdf below fc(e) → and D has the discrete pmf ( d=0 pp(d) = { }, d=1 l, d=1. Let X = C + D. Suppose you observe that X = . Find the conditional pmf pplx (d[x = 5).

Answer 1

4)

from the given Graph: fecc) =2c for CE [0,1] - to otherwise * Now, for a raudan lanable Crou) which is sum of a discrete and a continuous random Variable of livetoi mashion 4 The density function is; fx (x) = { fc (S-k). Pk KEK where K is the set of values which the discrete randam clarable (D) can take fx cau) a fe (1–0).+ fc (x-6)} Tove + fc (x-1); } f(x) = $ [fc Getfe (-43 +62(x+1)] for x = 5 we get fx (514) - cß fe (5/4)+fe (34)+FCC 417 Scanned with CamScanner

= § (0 + 2x + 2x + ] |fx( 547 = ?) arow's Poix Cdx=514) --P (d). f(x=54 1d) fx (slu) lets evaluate f (x = 5/41d) first Putting X=cto, we can write it as f(C+D=564/d) for d=0; f(c = 5/4)=0 d=tif (c= 3/4) = 2x 3 = 3/2 d=ţif (c=4) = 2x = Putting these values in the equation for the conditional pmf, we get; - Polx (d=0/x=5/4)=0 Poux (d = +/x=56) = $x - 34 2/3 de Pox (d=1\x=56) * xt Scanned with CamScanner

** It has been mentioned that the sum of a a discrete and a continuous rv is a continuous rv with the given pdf. In case you don't already know the result, here is the proof:

X =C+D

P(X <.) = P(C+D<x)= P(C<– D|D= k)P(D = k)

P(X 5x) = FC(x – k)px

Fc is the cumulative distribution function. Now differentiating it ,we get the PDF.

fx(x) =Σ fe(x – kype

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