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please I want to show that $F_n(t)=P{(S_n\le t})=\int_0^tF_{n-1}(t-x)f(x)dx.$

My problem here is I do not have enough hypothesis to work with. What I only know is that $F(t)=\int_0^tf(s)ds$ and $F(t)*f(t)=\int_0^tF(t-x)f(x)dx$.

I think induction will help, but I am not sure. I need some hint.

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What exactly is $S_n$? It seems to be the sum of $n$ independent random variables. Are these $n$ variables $n$ individual failure times with density function $f(s)$? And are you sure that $F_n(t)$ is a pdf which generally denotes a probability density function and not a cumulative probability distribution function which is often abbreviated as cdf or CDF? Your expression $F_n(t) = P(S_n \leq t)$ seems to indicate the latter. –  Dilip Sarwate Oct 19 '11 at 16:18
    
Yes, $S_n $ is the sum of $n$ independent random variables known as the time of the $nth$ failure. The density function of $F_n(t)$ is $f_n(t)$ for each n. –  Hassan Muhammad Oct 19 '11 at 16:30
    
"density function of $F_n(t)$"?? Are $f_1(t), f_2(t), \ldots, f_n(t)$ all the same function, e.g. $f_n(t) = \exp(-t), t > 0$ for all $n$, or are they different? And what about $f(x)$ appears in the first line of your question? Please revise your question to make it more easily understandable. –  Dilip Sarwate Oct 19 '11 at 16:35

1 Answer 1

up vote 1 down vote accepted

If $S_n=X_1+\cdots+X_n$ with $(X_n)$ independent and $X_n$ with density $f_n$, then $$ \mathrm P(S_n\leqslant t\mid X_n=x)=\mathrm P(S_{n-1}\leqslant t-x\mid X_n=x)=\mathrm P(S_{n-1}\leqslant t-x)=F_{n-1}(t-x), $$ where the second equality stems from the fact that $S_{n-1}$ and $X_n$ are independent. Integrating this over $x$, one gets $$ F_n(t)=\mathrm P(S_n\leqslant t)=\int\mathrm P(S_n\leqslant t\mid X_n=x)\mathrm dP_{X_n}(x)=\int F_{n-1}(t-x)f_n(x)\mathrm dx. $$ In the case at hand $S_0=0$ and $S_1=X_1$ hence one can start the recursion at $F_0(t)=\mathbf 1(t>0)$ or at $F_1(t)=\mathrm P(X_1\leqslant t)=\int\limits_0^tf(x)\mathrm dx$. And all the $X_k$ are almost surely positive hence $f_n(x)=0$ for $x\leqslant0$ and $F_{n-1}(t-x)=0$ for every $x\geqslant t$, thus the integral giving $F_n(t)$ may be restricted to $0\leqslant x\leqslant t$.

More generally, if $W=U+V$ for independent random variables $U$ and $V$ and if $U$ has CDF $G$ defined by $G(t)=\mathrm P(U\leqslant t)$ and $V$ has density $h$, then the CDF of $W$ is $$ \mathrm P(W\leqslant t)=\int G(t-x)h(x)\mathrm dx. $$

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Thanks Didier, I am happy with your help. –  Hassan Muhammad Oct 23 '11 at 21:48
    
Thanks. It is customary on the site to show one's appreciation of an answer by voting for it and, if one is the author of the question, by accepting the answer. –  Did Oct 23 '11 at 22:57

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