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yixinttt

ľ³æ (ÕýʽдÊÖ)

[ÇóÖú] LatexÁ½Ì춼û¸ã¶¨µÄ¹«Ê½

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LatexÁ½Ì춼û¸ã¶¨µÄ¹«Ê½
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yixinttt

ľ³æ (ÕýʽдÊÖ)

sorry£¬Íü¼ÇÌùÎҵĴúÂëÁË£¡

\begin{equation}
  \left\{
  \begin{aligned}
  \begin{array}{*{5}{c}}
  {P(X_{t+1}=d-1 \mid X_t=d)=
  \begin{cases}
  \frac {2^k-2^d}{2^k-1-t} & ($ 1 $<$ d+1 $ \le$ k $)\\
  0 & ($otherwise$)
  \end{cases}}\\
  {P(X_{t+1}=d \mid X_t=d)=
  \begin{cases}
  \frac {1}{2} \cdot \frac {2^d-1-t}{2^k-1-t} & ($ d $>$ 0, t+1 $\le$ $2^d-1$ $)\\
  0 & ($otherwise$)
  \end{cases}}\\
  P(X_{t+1}=0 \mid X_t=d) = P(X_{t+1} \mid X_t=d)  &{( d   >  0 )} \\
  p(X_{t+1} = 0 \mid X_t=0) = 1
  \end{array}
  \end{aligned}
   \right.
\end{equation}
Á˽âÁ˽â
2Â¥2015-06-13 22:31:41
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

intoabyss

½ð³æ (СÓÐÃûÆø)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

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yixinttt: ½ð±Ò+40, ²©Ñ§EPI+1, ¡ï¡ï¡ï¡ï¡ï×î¼Ñ´ð°¸, ½â¾öÁËÎÊÌ⣡ 2015-06-14 09:45:12
ÊÔÊÔ
\left\{ \begin{align}
  & P({{X}_{t+1}}=d-1\mid {{X}_{t}}=d)=\left\{ \begin{align}
  & \frac{{{2}^{k}}-{{2}^{d}}}{{{2}^{k}}-1-t},(1<d+1\le k), \\
& 0,otherwise; \\
\end{align} \right. \\
& P({{X}_{t+1}}=d\mid {{X}_{t}}=d)=\left\{ \begin{align}
  & \frac{1}{2}\cdot \frac{{{2}^{d}}-1-t}{{{2}^{k}}-1-t},(d>0,t+1\le {{2}^{d}}-1), \\
& 0,otherwise; \\
\end{align} \right. \\
& P({{X}_{t+1}}=0\mid {{X}_{t}}=d)=P({{X}_{t+1}}\mid {{X}_{t}}=d),(d>0); \\
& p({{X}_{t+1}}=0\mid {{X}_{t}}=0)=1 \\
\end{align} \right.
3Â¥2015-06-13 22:44:43
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

yixinttt

ľ³æ (ÕýʽдÊÖ)

ÒýÓûØÌû:
3Â¥: Originally posted by intoabyss at 2015-06-13 22:44:43
ÊÔÊÔ
\left\{ \begin{align}
  & P({{X}_{t+1}}=d-1\mid {{X}_{t}}=d)=\left\{ \begin{align}
  & \frac{{{2}^{k}}-{{2}^{d}}}{{{2}^{k}}-1-t},(1<d+1\le k), \\
& 0,otherwise; \\
\end{al ...

·Ç³£¸Ðл£¬¸ù¾ÝÄãµÄ´úÂëµ÷ÕûÁËÒ»ÏÂͨ¹ýÁË£¬ºÜ±ê×¼¡£Ð»Ð»À²
ÏÂÃæÊÇÎҸĺóµÄ£¬Æäʵ¾Í°Ñalign»»³ÉÁËaligned£¬Ö®Ç°×öµ½ÕâÒ»²½¾ÍÊǰÑalignedºÍleftλÖ÷ŷ´ÁË¡£

\begin{equation}
\left\{ \begin{aligned}
  & P({{X}_{t+1}}=d-1\mid {{X}_{t}}=d)=\left\{ \begin{aligned}
  & \frac{{{2}^{k}}-{{2}^{d}}}{{{2}^{k}}-1-t},(1<d+1\le k), \\
& 0,otherwise; \\
\end{aligned} \right. \\
& P({{X}_{t+1}}=d\mid {{X}_{t}}=d)=\left\{ \begin{aligned}
  & \frac{1}{2}\cdot \frac{{{2}^{d}}-1-t}{{{2}^{k}}-1-t},(d>0,t+1\le {{2}^{d}}-1), \\
& 0,otherwise; \\
\end{aligned} \right. \\
& P({{X}_{t+1}}=0\mid {{X}_{t}}=d)=P({{X}_{t+1}}\mid {{X}_{t}}=d),(d>0); \\
& p({{X}_{t+1}}=0\mid {{X}_{t}}=0)=1 \\
\end{aligned} \right.
\end{equation}
LatexÁ½Ì춼û¸ã¶¨µÄ¹«Ê½-2
¹«Ê½3.jpg

Á˽âÁ˽â
4Â¥2015-06-14 09:42:27
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
Ïà¹Ø°æ¿éÌø×ª ÎÒÒª¶©ÔÄÂ¥Ö÷ yixinttt µÄÖ÷Ìâ¸üÐÂ
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