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ankoles [38]
2 years ago
5

Express the following in the form of p/q where p and q are integers and q is not equal to 0.

Mathematics
2 answers:
ANEK [815]2 years ago
3 0

Answer:

1.06bar is not equal to zero

slava [35]2 years ago
3 0
The answer is 1 0.06
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If X is a r.v. such that E(X^n)=n! Find the m.g.f. of X,Mx(t). Also find the ch.f. of X,and from this deduce the distribution of
astraxan [27]
M_X(t)=\mathbb E(e^{Xt})
M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)
M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots
M_X(t)=1+t+t^2+t^3+\cdots
M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}

provided that |t|.

Similarly,

\varphi_X(t)=\mathbb E(e^{iXt})
\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots
\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)
\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)
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You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute

F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt

The integral can be rewritten as

\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt

so that

F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt

There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to x, which gives

\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt
=\displaystyle\frac{\pi(1-s)}{2s(1+s)}

and taking the inverse transform returns

F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}

which describes an exponential distribution with parameter \lambda=1.
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olchik [2.2K]

Answer: A

Step-by-step explanation:

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Leno4ka [110]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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