Question

Finding the distribution sum of a continuous and discrete random variable?

If S is a r.v with a gamma distribution with parameters \(\alpha, \beta\) and T is a r.v. with a Bernoulli(p) distribution, how can I find the distribution of S + T ? (S and T are independent)

Answer 1

@SithsAndGiggles @ganeshie8 @vishweshshrimali5 @kirbykirby

Answer 2

Sorry I am not in touch with Bernoulli's distribution for 2 years. :(

Answer 3

Does it help if I say the bernoulli distribution is
\(P(T=t)=p^t(1-p)^{1-t}, x\in \{0, 1\}\) ?

Answer 4

Gamma distribution ?

Answer 5

\(f_S(s)=\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}s^{\alpha-1}e^{-s/\beta}, s>0\) and the parameters \(\alpha>0, \beta>0\)

Answer 6

My initial guess was to try and use the CDF definition. Letting U = S+T, then\[P(U \le u)=P(U\le u|T=0)P(T=0)+P(U \le u|T=1)P(T=1)\] by the law of total probability, then
\[P(U \le u)=P(S+T\le u|T=0)(1-p)+P(S+T\le u|T=1)(p)\\
=P(S\le u)(1-p)+P(S+1\le u)\cdot p\\
P(S \le u)(1-p)+P(S\le u-1)\cdot p\]
I don't know if I can actually do that tho. I'm not that familiar with mixed distributions.

Answer 7

yes you can do that

Answer 8

Thank you!

Answer 9

@Zarkon If I want to show that it's a valid distribution, I want to show that it sums/integrates to 1. So, if I proceed by finding \(f_U(u)\), is it sufficient to say that \(f_U(u)= f_S(u)(1-p)+f_S(u-1)\cdot p\). Do I just take the derivative with respect to u? And how do I determine the support?

Answer 10

(Im assuming it's safe to take the derivative since S is the continuous r.v ? )

Answer 11

@ikram002p

Answer 12

to find the support of S+T you look at the support of S and T individually first

Since the support of S is \([0,\infty)\) and the support of T is [0,1] then the support of S+T is \([0,\infty)\)

to show it integrates to 1 just integrate it

\[\int\limits_0^{\infty}f_U(u)du=\int\limits_0^{\infty}( f_S(u)(1-p)+f_S(u-1)\cdot p)du\]


\[=\int\limits_0^{\infty}f_S(u)(1-p)du+\int\limits_0^{\infty}f_S(u-1)\cdot pdu\]

\[=(1-p)\int\limits_0^{\infty}f_S(u)du+p\int\limits_0^{\infty}f_S(u-1)du\]

\[=(1-p)\cdot 1+p\int\limits_0^{\infty}f_S(u-1)du\]

\[=(1-p)\cdot 1+p\left(\int\limits_0^{1}f_S(u-1)du+\int\limits_1^{\infty}f_S(u-1)\right)du\]

\[=(1-p)\cdot 1+p\left(0+1\right)du=(1-p)+p=1\]

since \(f_S(x)=0\) for \(x<0\)

Answer 13

I really appreciate help! Thanks very much